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3 years ago

How do I find the range of a function (no links)

Mathematics

1 answer:

Anna11 [10]3 years ago

6 0

D is your answer because the bar is 4 and you must have an equal amount of plates on each side so it cannot be B. It goes up by 40 because of 2 plates on each side of the bar

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Henry has 2 accounts with a credit limit of $6000. He has $500 in monthly debts, including a $200 car payment. He has never made

Volgvan

I think it's D as an answer but I could be wrong

8 0

3 years ago

Square ABCD and square EFGH share a common center on a coordinate plane. is parallel to diagonal . Across how many lines of refl

Oduvanchick [21]

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7 0

3 years ago

Read 2 more answers

Find the coefficient of the x^3 in the expansion of (2x-9)^5

il63 [147K]

Use the binomial theorem:

$\displaystyle (2x-9)^5 = \sum_{k=0}^5 \binom5k (2x)^{5-k}(-9)^k = \sum_{k=0}^5 \frac{5!}{k!(5-k)!} 2^5 \left(-\frac92\right)^k x^{5-k}$

The x ³ terms occurs for 5 - k = 3, or k = 2, and its coefficient would be

$\dfrac{5!}{2!(5-2)!} 2^5 \left(-\dfrac92\right)^2 = \boxed{6480}$

8 0

3 years ago

What is matrix X if 0.5X=A? Matrix A=[6 -12 4 6].

vladimir1956 [14]

Answer:

See attachment

Step-by-step explanation:

The given matrix equation is:

$0.5X = A.$

To find Matrix X, we need to multiply both sides of the equation by 2 to obtain:

$2 \times 0.5X =2 \times A$

This simplifies to;

$X = 2A$

By scalar multiplication, we multiply each entry in the matrix A by to 2 to obtain matrix X.

6 0

3 years ago

How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide

kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The margin of error of a (1 - α)% confidence interval for population mean (μ) is:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The information provided is:

σ = $60

MOE = $2

The critical value of z for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}$

$=[\frac{1.96\times 60}{2}]^{2}$

$=3457.44\\\approx 3458$

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0

3 years ago