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3 years ago

15

Which graph represents the solution of the system StartLayout Enlarged left-brace 1st row x squared + y squared = 4 2nd row x mi

nus y = 1 EndLayout?

1 answer:

svp [43]3 years ago

8 0

Answer:

B

Step-by-step explanation:

i put it in a graphing calculator and just took the test.

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Simplify the expression. 3x-9y-5x-7y

emmasim [6.3K]

3x - 9y - 5x - 7y

Combine like terms

( 3x + - 5x ) + ( - 9y + - 7y )

= - 2x - 16y

5 0

4 years ago

Read 2 more answers

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago

A triangle with an area of 3 mm² is dilated by a factor of 6. What is the area of the dilated triangle?

Contact [7]

When the sides are dilated by a factor of 6, the area increases by a factor of 6², which is 36.
3*36=108 mm²

8 0

3 years ago

The manufacturer of a certain type of new cell phone battery claims that the average life span of the batteries is 500 charges;

maxonik [38]

Answer:

a. 0

b. Yes

c. The manufacturers claim is not plausible

d. 0.3446

e. A sample mean life time of 39.8hr is not unusually short.

f. The manufacturers claim us plausible

Step-by-step explanation:

Please see attachment

8 0

4 years ago

A. f(1.3)=1. B. f(1.8)=2 c. f(-1.5)=-2 D. f(2.8)=2

dimulka [17.4K]

What’s the question?

7 0

3 years ago