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3 years ago

Which data set has the widest spread?

Mathematics

2 answers:

azamat3 years ago

6 0

C or D

widest spread is about a range whichever has lowest OR/AND highest number is correct

neonofarm [45]3 years ago

4 0

Answer:

Step-by-step explanation:

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10.)A child-sized swimming pool can be modeled by a cylinder. The pool has a diameter of 6 1/2 feet and a height of 12 inches. T

Luba_88 [7]

Answer:

165.47≈165 Gallons of Water

Step-by-step explanation:

According To the Question,

Given, A child-sized swimming pool can be modeled by a cylinder. The pool has a diameter of 6 1/2 feet (or) 6.5 feet (radius is 3.25 feet) and a height of 12 inches(1 feet).

The pool is filled with water to 2/3 of its height (H= 2/3 feet) .

Volume of Cylinder = π×R²×H , now put values, we get

Volume = 3.14×3.25²×2/3

Volume = 22.122 foot³ .

And, One cubic foot equals 7.48 gallons of water

Thus, 22.122 cubic foot equals to 165.47≈165 gallons of water .

8 0

2 years ago

A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se

bearhunter [10]

Answer:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

Step-by-step explanation:

We know the following info from the problem

$\bar X_1 = 14.5$ sample mean for the group 1

$s_1 = 3.98$ the standard deviation for the group 1

$n_1= 20$ the sample size for group 1

$\bar X_2 = 13.9$ sample mean for the group 2

$s_1 = 4.03$ the standard deviation for the group 2

$n_2= 17$ the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given :

$df = n_1 +n_2- 2 = 20+17-2=35$

The confidence level is 0.9 or 90% and the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 = 0.05$ and the critical value for this case is:

$t_{\alpha/2} = 1.69$

And replacing the info given we got:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

5 0

2 years ago

The value of the 8 in 89.945 is how many times rhe value of the 8 in 763.186

trasher [3.6K]

Answer:

1000x

Step-by-step explanation:

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3 years ago

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CaHeK987 [17]

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