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3 years ago

PLEASE help with these 2 questions thank you!

Mathematics

1 answer:

Sergio039 [100]3 years ago

6 0

Download the app photo math help a whole lot

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Mr. Sanchez buys 5 shirts and 3 ties for $34. At

Nastasia [14]

Answer:

y < x + 7

Step-by-step explanation:

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3 years ago

What is the measure of angle ABC

zvonat [6]

ANSWER = c. 63°

EXP:

>ABC = 1/2(x + y)

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3 years ago

: (2x – Зу = 18 (2x + y = 2

charle [14.2K]

Answer:

2x = 2 × 12 = 24 24 =2x 24-6=18

3y = 3 × 2 = 6 6 =3y

2x = 2×1 = 2 2 =2x 2+0=2

y = 0 0 = y

Step-by-step explanation:

I'm not sure how the question is mentioned in the test but from my point of view I got this not sure how the it is explained in your nest im only 11 so yeh. sos if its wrong

7 0

3 years ago

Read 2 more answers

What is the slope of the line passing through (-3,5) and (5,-3)?

Anastasy [175]

Answer:

The Answer is: Slope = -1

Step-by-step explanation:

Given points: (-3, 5) and (5, -3).

Use the equation for slope, m:

m = y - y1/(x - x1)

m = 5 - (-3)/(-3 - 5)

m = 5 + 3/-8

m = 8/-8 = -1

Hope this helps! Have an Awesome Day!! :-)

4 0

3 years ago

Read 2 more answers

Assuming that s = 18,000 is a reasonable estimate for what sample size would be needed to ensure that we could estimate the true

djverab [1.8K]

Answer:

The minimum sample size required is 71.

Step-by-step explanation:

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

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

The margin of error for this interval is:

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

It is provided that the population standard deviation σ can be estimated by the sample standard deviation s.

The given information is:

s = 18000,

Margin of error = $4200.

Confidence level = 95%

The z-value for 95% confidence level is,

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

*Use a z-table for the critical value.

Compute the sample size required as follows:

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

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

$=[\frac{18000\times 1.96}{4200}]^{2}$

$=70.56\\\approx71$

Thus, the minimum sample size required is 71.

3 0

3 years ago