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2 years ago

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ME in the coordinate plane has endpoints with coordinates (1,4) and (−10,−4). Graph ME and find two possible locations for point

R, so R divides ME into two parts with lengths in a ratio of 2:3

1 answer:

Solnce55 [7]2 years ago

8 0

Answer:

the answer is on google

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The measure of angle 4 is 45. If angle 4 is congruent to angle 1, what is the measure of angle 5?

mr_godi [17]

The answer is A) 45°. This is because angles 1 and 5 are vertical angles. Vertical angles are congruent.

8 0

3 years ago

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

(3 − 4i)(5 + 2i)<br><br> A. 15 − 8i<br> B. 23 − 14i<br> C. 15 − 8i2<br> D. 15 − 14i − 8i2

choli [55]

Answer:

B.23-14i

Step-by-step explanation:

(3x5)+(3x2i)-(4ix5)-(4ix2i)

15+6i-20i-8(-1)

15+6i-20i+8

23-14iฟ

8 0

3 years ago

Marine scientists categorize signature whistles of bottlenose dolphins by typelong dash—type a, type b, type c, etc. In one s

r-ruslan [8.4K]

Answer:

The 95% confidence interval for the true proportion of bottlenose dolphin signature whistles that are type a whistles is (0.4687, 0.6123).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 185, \pi = \frac{100}{185} = 0.5405$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5405 - 1.96\sqrt{\frac{0.5405*0.4595}{185}} = 0.4687$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5405 + 1.96\sqrt{\frac{0.5405*0.4595}{185}} = 0.6123$

The 95% confidence interval for the true proportion of bottlenose dolphin signature whistles that are type a whistles is (0.4687, 0.6123).

3 0

3 years ago

What is -35,840-(-14,764)

earnstyle [38]

21076 is the answer

it's like saying -35840 + 14764

3 0

2 years ago