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3 years ago

Can u guys please answer these questions

Mathematics

1 answer:

aliya0001 [1]3 years ago

8 0

Answer:

Im not sure about Q1 but I'll help with Q2

9 =

$\sqrt{81}$

8 =

$\sqrt[3]{512}$

14 =

$\sqrt{196}$

2 =

$\sqrt[3]{8}$

4 =

$\sqrt{16}$

I hope this helps :)

If you would like me to, i will respect your wishes and delete my answer.

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Write a real world problem that you would represent with the equation 4x+5=37.

AURORKA [14]

Bob was 5 feet off the ground. He travels 4 feet every single second. How long will it take for him to reach 37 feet?

4x + 5 = 37

4x + 5 (-5) = 37 (-5)

4x = 32

4x/4 = 32/4

x = 8

It will take Bob ~8 seconds to reach 37 feet*

*Problem & answer

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13- 0.7w + 8x when w = 12 and x = 1/2

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Simplify the expression and you get: 8.6
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Can you help me figure out the slope of this line? please and thx ASAP.

KonstantinChe [14]

The rise is the vertical distance between the two points, which is the difference between their y-coordinates. That makes the rise y2 − y1. The run between these two points is the difference in the x-coordinates, or x2 − x1.

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3 years ago

PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST

ohaa [14]

Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

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erastova [34]

Answer:38

Step-by-step explanation:

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