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If 48 feet are represented by 12 inches on a scale drawing, then a scale of inch represents how many feet?

deff fn [24]

1 inch represents 4 feet

6 0

3 years ago

A can of tuna is in the shape of a cylinder. The distance around the top of the cylinder is approximately 251.2 mm. The tuna com

bulgar [2K]

Answer: 39.9797217047

Step-by-step explanation:

2πr = 251.2

πr = 125.6

r = 39.9797217047

5 0

2 years ago

Read 2 more answers

26=4v+3v+12 can anyone help?

motikmotik

Answer:

Step-by-step explanation:

Combine like terms

3v + 4v

26=7v+12

Subtract 12 from both sides of the equation

26-12

12-12

14=7v

Simplify

so subtract the 12's

Divide both sides of the equation by the same term

14 divided by 7

7 divided by 7

Simplify

do the division and move the term on the other side and it would be

v=2

Solution

v=2

4 0

3 years ago

Read 2 more answers

Evaluate the function rule to find the range of each function for the domain {-3, 0, 5}. f(x) = x² − 6x + 4

harina [27]

Range is output for given input

domain is D={-3,0,5}
input -3,0 and 5 for x

f(-3)=(-3)^2-6(-3)+4
f(-3)=9+18+4
f(-3)=31

F(0)=(0)^2-6(0)+4

f(0)=0-0+4

f(0)=4

f(5)=(5)^2-6(5)+4

f(5)=25-30+4

f(5)=-1

Domain={-3,0,5}

Range={31,4,-1}

5 0

3 years ago

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

Yuliya22 [10]

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

We can calculate the mean and deviation for eaach group with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =5.92$ represent the sample mean for the group 1

$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

And now we can calculate the statistic:

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

The degrees of freedom are given by:

$df=5+5-2=8$

And now we can calculate the p value using the altenative hypothesis:

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

7 0

3 years ago