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3 years ago

50 POINTS IF YOU ANSWER CORRECTLY!

Mathematics

2 answers:

Katena32 [7]3 years ago

8 0

Answer:

wow

Step-by-step explanation:

melisa1 [442]3 years ago

4 0

Answer:

Problem 4: Option C, 32

Problem 5: You need to have the 2 and the x combine. Make the table 2x than 16.

Problem 6: x = 17

Problem 7: She drew a 2, x and an 16. Instead you must have 2x and 16. 2x + 16 = 50

Step-by-step explanation:

Problem 4

2 + x + 16 = 50

x + 18 - 18 = 50 - 18

x = 32

Answer: Option C, 32

Problem 5

You need to have the 2 and the x combine. Make the table 2x than 16.

Answer: You need to have the 2 and the x combine. Make the table 2x than 16.

Problem 6

2x + 16 - 16 = 50 - 16

2x / 2 = 34 / 2

x = 17

Answer: x = 17

Problem 7

She drew a 2, x and an 16. Instead you must have 2x and 16. 2x + 16 = 50

Answer: She drew a 2, x and an 16. Instead you must have 2x and 16. 2x + 16 = 50

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Step-by-step explanation:

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write the equation of the line in slope intercept form with the given information: through the points (-1,-9) and (2,6) how woul

Anit [1.1K]

Slope intercept form isy=mx+b
m=slope
b=y intercept
ok so to find the slope when given 2 points you do
slope=(y1-y2)/(x1-x2)

(x,y)
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(2,6)
x1=-1
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(2,6)
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3 years ago

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (

alisha [4.7K]

Answer:

Point estimation of the mean = 17.6

Error = 4.6

95% CI: 13.0 ≤ μ ≤ 22.2

The distribution can be approximated to a normal because the sample size n=50 is bigger than 30.

Step-by-step explanation:

We have this information:

- Sample mean:

$\bar{X}=17.598$

- Sample standard deviation

$s=16.01712719$

- Sample size

$n=50$

For now on, we round to one decimal place.

The best estimation for the population mean is the sample mean

$\mu=\bar{X}=17.6$

The margin of error is equal to the t-value multiplied by the sample standard deviation and divided by the sample size.

The t value depends on the degrees of freedom and the width of the confidence interval. In this case it is a 95% CI and the degrees of freedom are 49. For this conditions, the t-value is t=2.01.

Then, the margin of error is

$E=t_{49}*\sigma/\sqrt{n}=2.01*16.0/\sqrt{50}=4.553$

Then, the confidence interval can be constructed as:

$\bar{X}-t\sigma/\sqrt{n}\leq\mu\leq\bar{X}-t\sigma/\sqrt{n}\\\\17.6-4.6\leq \mu \leq 17.6+4.6\\\\ 13.0\leq \mu \leq 22.2$

7 0

3 years ago