Crickets and math <br> Thanks for help

The length of a rectangle is 12 ft. longer than twice its width. If the total length of a width and a length is 44 ft., what is

REY [17]

Width of the rectangle is 10.67 ft

<u>Step-by-step explanation:</u>

Step 1:

Let the width of the rectangle be x. Then length = 2x + 12

Given width + length = 44ft ⇒ x + 2x + 12 = 44

⇒ 44 = 3x + 12

⇒ 32 = 3x

⇒ x = 32/3 = 10.67

Width of the rectangle = 10.67 ft

7 0

3 years ago

What value of k will make this expression equal to 0? <br><br> b • 3.8 • 2.72 • (k – 4)

g100num [7]

4-4=0

anything x 0=0

your answer is 4

Hope this helps!

5 0

3 years ago

Read 2 more answers

What is the final answer?

MakcuM [25]

<h3>Answer: 42</h3>

Explanation:

We have y = -0.9x^2 + 76x - 250 which is in the form y = ax^2+bx+c

where,

a = -0.9
b = 76
c = -250

The vertex (h,k) is when the profit is maxed out.

h = -b/(2a)

h = -76/(2(-0.9))

h = 42.222 approximately

Let's plug in x values around x = 42

Try x = 41

y = -0.9x^2 + 76x - 250

y = -0.9(41)^2 + 76(41) - 250

y = 1353.10

Now try x = 42

y = -0.9x^2 + 76x - 250

y = -0.9(42)^2 + 76(42) - 250

y = 1354.4

Now try x = 43

y = -0.9x^2 + 76x - 250

y = -0.9(43)^2 + 76(43) - 250

y = 1353.9

We see that the largest profit happens when x = 42.

3 0

2 years ago

(2k+11)<br> 131°<br> What is the value of k

Triss [41]

The value of k is 60

131 - 11 = 120
120/2 = 60

hope this helped !

6 0

3 years ago

Read 2 more answers

For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les

kaheart [24]

Answer:

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

5 0

2 years ago

Crickets and math Thanks for help