I need help with these points+brainliest

(B) 0.9%. We can use the formula for compound interest to find the value of the investment after 2 years:

A = P(1 + r/n)^(nt)

where A is the amount of money after the time period, P is the principal (initial investment), r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the time in years.

Plugging in the given values, we get:

A = 1000(1 + 0.018/2)^(2*2)

= 1000(1 + 0.009)^4

= 1000(1.009)^4

≈ 1083.02

So the investment is worth approximately $1083.02 after 2 years.

To find the interest rate per year, we can use the formula:

r = n[(A/P)^(1/nt) - 1]

Plugging in the values we know, we get:

r = 2[(1083.02/1000)^(1/(2*2)) - 1]

= 2[(1.08302)^(1/4) - 1]

≈ 0.9%

Therefore, the answer is (B) 0.9%.

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The future value of $2000 earning 18% interest, compounded monthly, for 4 years is $6,116.23.

To calculate the future value, we use the formula:

[tex]FV = P(1 + r/n)^{nt}[/tex]

Where:

FV is the future value

P is the principal amount ($2000)

r is the annual interest rate (18%)

n is the number of times the interest is compounded per year (12 for monthly compounding)

t is the number of years (4)

Plugging in these values, we get:

[tex]FV = 2000(1 + 0.18/12)^{12*4}[/tex]

FV = 6116.23

Therefore, the future value of $2000 earning 18% interest compounded monthly for 4 years is approximately $6,116.23. This means that if you invest $2000 today and earn 18% interest compounded monthly for 4 years, your investment will grow to $6,116.23 at the end of the 4-year period. It's important to note that the actual value may vary depending on the exact compounding method used by the bank or financial institution.

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The we have shown that S is a subspace of Rn and that x + y ∈ S.

Suppose ~u1, ~u2, . . . , ~uk∈Rnand let S = span( ~u1, ~u2, . . . , ~uk). Suppose ~x, ~y ∈S.(d) Show that ~x + ~y ∈S.The solution is given as follows.The problem statement is given below.Suppose u1, u2, . . . , uk∈Rnand let S = span(u1, u2, . . . , uk). Suppose x, y ∈S. Show that x + y ∈S.To prove that x + y ∈ S, we must first prove that S is a subspace of Rn. We can then show that x + y is a linear combination of vectors in S, so it must be in S.Let's get started with the proof.Let S = span{u1, u2, . . . , uk} be a subspace of Rn. Then u1, u2, . . . , uk are linearly independent, which implies that they span S. Thus, any vector in S can be written as a linear combination of u1, u2, . . . , uk.Suppose x, y ∈ S. Then x = a1u1 + a2u2 + . . . + akuk, and y = b1u1 + b2u2 + . . . + bkuk, for some scalars a1, a2, . . . , ak and b1, b2, . . . , bk.Now we can show that x + y is a linear combination of vectors in S, so it must be in S.x + y = (a1 + b1)u1 + (a2 + b2)u2 + . . . + (ak + bk)ukSince a1, a2, . . . , ak and b1, b2, . . . , bk are scalars, (a1 + b1), (a2 + b2), . . . , (ak + bk) are also scalars. Thus, x + y is a linear combination of u1, u2, . . . , uk, and hence x + y ∈ S. Therefore, we have shown that S is a subspace of Rn and that x + y ∈ S.

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After considering the speeds of the student and the teacher and the distance they ran in each case, we found that the students wins in all the races.

Speed is defined as the ratio of distance travelled to the amount of time it took. As speed simply has a direction and no magnitude, it is a scalar quantity.

An object is considered to be moving at a uniform speed when it travels the same distance in the same amount of time.

When an object travels a different distance at regular intervals, it is said to have variable speed.

Average speed is the constant speed determined by the ratio of the total distance travelled by an object to the total amount of time it took to travel that distance.

Given,

The length of the race = 100  yards

1st race

The speed of student = 10 yards/s

The speed of teacher = 3.75 yards / s

Time taken by student = distance / speed = 100/10 = 10s

Time taken by teacher = 100 / 3.75 = 26.67 s

So the student finishes the race because he/she took less time.

2nd race

Distance run by teacher = 100 - 10 = 90

Speed of teacher =  3.75 yards / s

Time taken by teacher = 90/3.75 =  24s

The distance and speed of the student are the same.

Time taken by student = 10 s

Still, the student finishes the race first.

3rd race

Distance run by teacher = 100 - 10 = 90

Speed of teacher =  3.75 * 2 yards /s = 7.5 yards/s

Time taken by teacher = 90/ 7.5 =  12s

The distance and speed of the student are the same.

Time taken by student = 10 s

Still, the student finishes the race first.

Therefore by considering the speeds of the student and the teacher and the distance they ran in each case, we found that the students wins in all the races.

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The expected rabbit population by next April is 486.

The rabbit population in a forest area can be found using the formula y=200(2.7)^0.08t, where y is the expected population, t is the time in months, and 200 is the initial population. To find the expected population by next April, we need to plug in the value of t as 12 (since there are 12 months in a year) and solve for y.

y=200(2.7)^0.08t

y=200(2.7)^0.08(12)

y=200(2.7)^0.96

y=200(2.4315)

y=486.3

Since the question asks for the expected population rounded to the nearest whole number, we can round 486.3 to 486.

Therefore, the expected rabbit population by next April is 486.

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The solution is  \boxed{A^{-1} = \left[\begin{array}{rr}-\frac{1}{3} & 0 \\ 0 & -\frac{1}{5}\end{array}\right]} and \boxed{A^{-1} = \left[\begin{array}{rrr}-\frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{12} & 0 \\ 0 & 0 & -\frac{1}{8}\end{array}\right]}.

(a) To find the inverse of \( A=\left[\begin{array}{rr}-3 & 0 \\ 0 & 5\end{array}\right] \), we need to use the formula:

\( A^{-1} = \frac{1}{ad-bc} \left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right] \)

where \( a=-3, b=0, c=0, d=5 \).

Plugging in the values, we get:

\( A^{-1} = \frac{1}{(-3)(5)-(0)(0)} \left[\begin{array}{rr}5 & 0 \\ 0 & -3\end{array}\right] \)

\( A^{-1} = \frac{1}{-15} \left[\begin{array}{rr}5 & 0 \\ 0 & -3\end{array}\right] \)

\( A^{-1} = \left[\begin{array}{rr}-\frac{1}{3} & 0 \\ 0 & -\frac{1}{5}\end{array}\right] \)

So, the inverse of \( A=\left[\begin{array}{rr}-3 & 0 \\ 0 & 5\end{array}\right] \) is \( A^{-1} = \left[\begin{array}{rr}-\frac{1}{3} & 0 \\ 0 & -\frac{1}{5}\end{array}\right] \).

(b) To find the inverse of \( A=\left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{arra \), we need to use the formula:

\( A^{-1} = \frac{1}{\det(A)} \left[\begin{array}{rrr}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]^{T} \)

where \( a_{11}=4, a_{12}=0, a_{13}=0, a_{21}=0, a_{22}=-2, a_{23}=0, a_{31}=0, a_{32}=0, a_{33}=3 \) and \( \det(A) = (4)(-2)(3) - (0)(0)(0) - (0)(0)(0) = -24 \).

Plugging in the values, we get:

\( A^{-1} = \frac{1}{-24} \left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{array}\right]^{T} \)

\( A^{-1} = \frac{1}{-24} \left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{array}\right] \)

\( A^{-1} = \left[\begin{array}{rrr}-\frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{12} & 0 \\ 0 & 0 & -\frac{1}{8}\end{array}\right] \)

So, the inverse of \( A=\left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{arra \) is \( A^{-1} = \left[\begin{array}{rrr}-\frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{12} & 0 \\ 0 & 0 & -\frac{1}{8}\end{array}\right] \).

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Answer:

84

Step-by-step explanation:

6.3+16.1=22.4

22.4/2=11.2

11.2x7.5=84

Answer:

78.12 cm^2

Step-by-step explanation:

i will call the 6.3 cm line A, 7.5 line B and 16.1 cm C

first we divide this into two shapes, a rectangle and a triangle

THE RECTANGLE:

the formula to find the area of a rectangle: L x W

A x B

6.3 x 7.5 = 47.25 cm^2

THE TRIANGLE:

this is a right angle triangle

formula to find the area of a right angle triangle: (1/2) x BASE x H

B is parallel to H and share the same length

to find BASE we need to subtract 6.3cm from 16.1cm (check attachment)

16.1 - 6.3 = 9.8

(1/2) x 9.8 x 6.3 = 30.87 cm^2

FINAL VALUE:

now we simply add our two values of area together

30.87 + 47.25 = 78.12 cm^2

In the word problem, The account have left $67 in 13 weeks.

Word problems are often described verbally as instances where a problem exists and one or more questions are posed, the solutions to which can be found by applying mathematical operations to the numerical information provided in the problem statement. Determining whether two provided statements are equal with respect to a collection of rewritings is known as a word problem in computational mathematics.

Here the given expression represents the medical savings account.

=> y=-24x+379

Here y represent  of money and x represent number of weeks.                                                                                                                            

Here Amount of money y = $67 then,

=> 67=- 24x+ 379

=> -24x = 67-379

=> -24x= -312

=> x = -312/-24

=> x = 13

Hence The account have left $67 in 13 weeks.

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Using Pivotal Condensation, the determinant of matrix A is 5.

Step 1: We start by initializing D as 1.

Step 2: We use the first row for pivotal condensation.
Row 0:  0 * D + 2 * 1 + 2 * 0 = 0
Row 1:  1 * D + 0 * 1 + 3 * 0 = 1
Row 2:  2 * D + 1 * 1 + 1 * 0 = 2

Step 3: We make the first row entries 0 by multiplying the entire row by (-2).
Row 0: -0 * D - 2 * 1 - 2 * 0 = 0
Row 1:  1 * D + 0 * 1 + 3 * 0 = 1
Row 2:  2 * D + 1 * 1 + 1 * 0 = 2

Step 4: We add row 0 to row 1 and row 0 to row 2.
Row 0:  0 * D + 2 * 1 + 2 * 0 = 0
Row 1:  1 * D + 0 * 1 + 3 * 0 = 1
Row 2:  0 * D + 3 * 1 + 3 * 0 = 3

Step 5: We make the entries of the second row 0 by multiplying the entire row by (-1/3).
Row 0:  0 * D + 2 * 1 + 2 * 0 = 0
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  0 * D + 3 * 1 + 3 * 0 = 3

Step 6: We add row 1 to row 0 and row 1 to row 2.
Row 0:  1/3 * D + 2 * 1 + 2 * 0 = 1/3
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  3/3 * D + 3 * 1 + 3 * 0 = 5

Step 7: We multiply the entries of the first row by (-3) to make the entries of the first row 0.
Row 0:  0 * D + 6 * 1 + 6 * 0 = 0
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  3/3 * D + 3 * 1 + 3 * 0 = 5

Step 8: We multiply the last row by D.
Row 0:  0 * D + 6 * 1 + 6 * 0 = 0
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  5 * D + 3 * 1 + 3 * 0 = 5D

Step 9: We subtract row 1 from row 0 and row 1 from row 2.
Row 0:  4/3 * D + 6 * 1 + 6 * 0 = 4/3D
Row 1: -1/3 * D - 0 * 1 - 3 * 0 = -1/3
Row 2:  4/3 * D + 3 * 1 + 3 * 0 = 4/3D

Step 10: We calculate the determinant by multiplying the last row entries.

Determinant of matrix A is 5.

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The value of the function (f·g)(-9) is 186.

Functions are the fundamental part of the calculus in mathematics. The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input x.

The given functions f(x)=2x²-4x-15 and g(x)=x+12.

Here, (f·g)(x)=f(x)+g(x)

= 2x²-4x-15+x+12

= 2x²-3x-3

(f·g)(-9)=2(-9)²-3(-9)-3

= 2×81+27-3

= 162+27-3

= 162+24

= 186

Therefore, the value of (f·g)(-9) is 186.

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To determine whether a given vector is a solution to the equation \(A\vec{x}=\vec{0}\), we need to multiply the matrix A with the vector \(\vec{x}\) and check if the resulting vector is equal to the zero vector \(\vec{0}\).

Let's consider the first vector \(\vec{x_1}=\left[\begin{array}{c} 2 \\ 1 \\ 1 \end{array}\right]\). Multiplying the matrix A with this vector, we get:
\[ A\vec{x_1}=\left[\begin{array}{ccc} 3 & 6 & -6 \\ -1 & -2 & 5 \\ -1 & -2 & 2 \\ -1 & -1 & 0 \end{array}\right] \left[\begin{array}{c} 2 \\ 1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 12 \\ -4 \\ -3 \\ -3 \end{array}\right] \]
Since the resulting vector is not equal to the zero vector \(\vec{0}\), the vector \(\vec{x_1}\) is not a solution to the equation \(A\vec{x}=\vec{0}\).
Similarly, we can check for the other vectors:
For the vector \(\vec{x_2}=\left[\begin{array}{c} -1 \\ 2 \\ -1 \end{array}\right]\):
\[ A\vec{x_2}=\left[\begin{array}{ccc} 3 & 6 & -6 \\ -1 & -2 & 5 \\ -1 & -2 & 2 \\ -1 & -1 & 0 \end{array}\right] \left[\begin{array}{c} -1 \\ 2 \\ -1 \end{array}\right]=\left[\begin{array}{c} 6 \\ 1 \\ 1 \\ 1 \end{array}\right] \]

Again, the resulting vector is not equal to the zero vector \(\vec{0}\), so the vector \(\vec{x_2}\) is not a solution to the equation \(A\vec{x}=\vec{0}\).
For the vector \(\vec{x_3}=\left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right]\):
\[ A\vec{x_3}=\left[\begin{array}{ccc} 3 & 6 & -6 \\ -1 & -2 & 5 \\ -1 & -2 & 2 \\ -1 & -1 & 0 \end{array}\right] \left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \]
In this case, the resulting vector is equal to the zero vector \(\vec{0}\), so the vector \(\vec{x_3}\) is a solution to the equation \(A\vec{x}=\vec{0}\).

Therefore, only the vector \(\vec{x_3}=\left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right]\) is a solution to the equation \(A\vec{x}=\vec{0}\).

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Answer:

Step-by-step explanation:

20

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

multiply x by y multiply x by y multiply x by y 3 plus 5 minus 5 to the power of 2 then divide 3 plus 87 then you give your mom and dad a high five then go make out with a girl for an hour and then your answer is 72 to the power of x multiplyed by y.

Answer: ??

Step-by-step explanation:

The rate at which the petrol was used in kilometers per liter is 10 km/L that means that for every litre of petrol used for driving, Peter's car could travel 10 kilometers.

Peter drove a thousand km and used 100 liters of petrol. To find the rate at which the petrol was utilized in kilometers consistent with liter, we divide the whole distance traveled via the quantity of petrol used.

The formulation for this is rate of petrol usage = overall distance traveled/quantity of petrol used.

Rate of petrol usage = total distance traveled/quantity of petrol used

Substituting the values we get,

= 1000 km / 100 L

= 10 km/L

This means that for every liter of petrol used, Peter was capable of travel 10 kilometers.

Thus, the rate at which the petrol was used in kilometers per liter is 10 km/L.

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B

B=G-11 and B+G=28 could be used to solve the problem

Out of 300 grams, 147 grams of food is not water in the given food sample.

If 51% of a certain food is water, then 49% of it must be something other than water. To find out how many grams of the food are not water, we can first calculate how many grams of water there are,

300 grams x 51% = 153 grams of water

Then we can subtract this from the total weight to find the amount of food that is not water -

300 grams - 153 grams = 147 grams of food that is not water

Therefore, out of 300 grams of nutrition, 153 grams are water and 147 grams are not.

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Step-by-step explanation:

always remember Pythagoras for life :

c² = a² + b³

c is the Hypotenuse (the longest side of the right-angled triangle, it is opposite of the 90° angle). a and b are the legs.

c² = 19² + 17² = 361 + 289 = 650

c = sqrt(650) = 25.49509757... in ≈ 25.5 in

the Hypotenuse is about 25.5 in long.

Therefore, the probability of Blake making more than 7 out of 25 shots in practice is 0.9984.

To determine the probability of Blake making more than 7 out of 25 shots in practice, we can use the binomial probability formula: P(X = x) = (n choose x) * p^x * (1-p)^(n-x), where n is the number of trials, x is the number of successes, and p is the probability of success.

In this case, n = 25, p = 0.52, and we want to find the probability of x > 7. We can calculate this by finding the probability of x ≤ 7 and subtracting it from 1.

P(X > 7) = 1 - P(X ≤ 7)

= 1 - [(25 choose 0) * (0.52)^0 * (0.48)^25 + (25 choose 1) * (0.52)^1 * (0.48)^24 + ... + (25 choose 7) * (0.52)^7 * (0.48)^18]

= 1 - 0.0016

= 0.9984

Therefore, the probability of Blake making more than 7 out of 25 shots in practice is 0.9984.

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To complete the frequency table and ogive for the number of heads flipped, we need to determine the frequency of each of the indicated intervals. We can do this by counting the number of times each interval appears in the data set and entering the frequencies into the table.

Here is how to do it step-by-step:
1. Start with the first interval, which is 0-4 heads. Count the number of times this interval appears in the data set. For example, if there are 3 occurrences of 0-4 heads, enter 3 in the frequency column for this interval.
2. Repeat this process for each of the remaining intervals, counting the number of occurrences and entering the frequencies into the table.
3. Once you have entered all the frequencies, you can create the ogive by plotting the cumulative frequencies on a graph. Start with the first interval and plot the cumulative frequency at the upper limit of the interval. For example, if the first interval is 0-4 heads and the frequency is 3, plot the point (4,3) on the graph.
4. Continue this process for each of the remaining intervals, adding the frequency of each interval to the cumulative frequency and plotting the point at the upper limit of the interval.
5. Once you have plotted all the points, connect them with a line to create the ogive.

Here is the completed frequency table and ogive for the number of heads flipped:
| Interval | Frequency |
|----------|-----------|
| 0-4      | 3         |
| 5-9      | 5         |
| 10-14    | 4         |
| 15-19    | 2         |
| 20-24    | 1         |

Ogive:
(4,3) (9,8) (14,12) (19,14) (24,15)

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68% of the races he competed in had a finish time around 64.5 and 65.5 seconds.

The term "variance" (or "") refers to an assessment of the data's dispersion from the mean. A small variance implies that the data are grouped around the normal, and while a large standard deviation shows that the data are more dispersed.

[tex]\begin{aligned}& \mathrm{P}(\mu-\sigma < \mathrm{X} < \mu+\sigma) \approx 68 \% \\& \mathrm{P}(\mu-2 \sigma < \mathrm{X} < \mu+2 \sigma) \approx 95 \% \\& \mathrm{P}(\mu-3 \sigma < \mathrm{X} < \mu+3 \sigma) \approx 99.7 \%\end{aligned}[/tex]

When the standard deviation out from mean of the distribution of X is and the mean of the dispersion of X is (assuming X is normally distributed).

Kiran's 400-meter dash timings have an average of 65 secs and a confidence interval of 0.5 seconds, and they are regularly distributed.

Using the formula, we then obtain

[tex]\mathrm{P}(65-0.5 < \mathrm{X} < 65+0.5)=\mathrm{P}(64.5 < \mathrm{X} < 65.5) \approx 68 \%[/tex]

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Answer:

19/3

Step-by-step explanation:

Peter ended up with 31 beads, and Dan ended up with 28 beads.

What is the fraction?

A fraction is a mathematical expression that represents a part of a whole. It is written in the form of a ratio between two numbers, with the top number called the numerator and the bottom number called the denominator.

Let's represent the number of beads that Peter and Dan had at the start by P and D, respectively. Then we can set up an equation based on the information given in the problem:

After giving away 1/4 of his beads, Peter had 3/4 of his original number of beads, which is (3/4)P.

After giving away 1/5 of his beads, Dan had 4/5 of his original number of beads, which is (4/5)D.

According to the problem, both had the same number of beads left after giving away some of their beads:

(3/4)P = (4/5)D

We also know that Peter had 7 more beads than Dan at the start:

P = D + 7

We can use substitution to solve for D:

(3/4)(D+7) = (4/5)D

9D/20 + 21/20 = 4D/5

D = 35

So Dan had 35 beads at the start. Using the equation P = D + 7, we can find that Peter had:

P = D + 7 = 35 + 7 = 42

After giving away 1/4 of his beads, Peter had (3/4)P = (3/4)*42 = 31.5 beads, which we can round down to 31 beads since we're dealing with whole numbers of beads. After giving away 1/5 of his beads, Dan had (4/5)D = (4/5)*35 = 28 beads.

Therefore, Peter ended up with 31 beads, and Dan ended up with 28 beads.

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The area of the resulting shape is 90 sq meters

The cylinder represents the given parameter

Such that we have the following dimensions

Radius = 6 cm

Height = 15 cm

When the solid cylinder is cut vertically through its center, we have a rectangle with the following dimensions

Length = 6 cm

Width = 15 cm

The area is then calculated as

Area = Length * Width

So, we have

Area = 6 * 15

Evaluate

Area = 90

Hence, the area is 90 square meters

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Answer:

c

Step-by-step explanation:

it makes sense

Answer:

Answer:

Linear equation to represent the total cost:

Let x be the number of kgs of tomatoes Mail bought, and y be the number of kgs of potatoes Mail bought. The cost of x kgs of tomato at Rs. 50 per kg is 50x, and the cost of y kgs of potato at Rs. 20 per kg is 20y. Therefore, the total cost of the vegetables is:

Total cost = 50x + 20y

Substituting the value of total cost as Rs. 200, we get:

50x + 20y = 200

This is the required linear equation to represent the total cost.

Finding the value of x:

Let's assume that Mail bought x kgs of tomato and 2.5 kgs of potato. Using the equation derived above:

50x + 20(2.5) = 200

Simplifying the equation:

50x + 50 = 200

50x = 150

x = 3

Therefore, Mail bought 3 kgs of tomato.

Finding the point at which the graph of 5x+2y=20 cuts x-axis:

To find the point at which the graph of 5x+2y=20 cuts the x-axis, we need to set y = 0 in the equation and solve for x:

5x + 2(0) = 20

5x = 20

x = 4

Therefore, the point where the graph of 5x+2y=20 cuts the x-axis is (4,0).

Answer:

4-2-5

Step-by-step explanation:

16, 8, and 20 can all be divided by 4

Leaving you with 4 2 and 5.

The ratio is 4 to 2 to 5

Step-by-step explanation: