Find the missing lengths m and n in the rectangular prism.

A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground directly under the cable and x feet

gayaneshka [121]

Answer:

293.38 pounds

Step-by-step explanation:

We are given that

Distance between poles=35 feet

$h(x)=10+0.1(x^{1.5})$

Weight of cable=10.4 per linear foot

We have to find the weight of the cable.

Differentiate w.r.t

$h'(x)=0.1(1.5)x^{0.5}=0.15x^{0.5}$

$s=2\int_{0}^{17.5}\sqrt{1+(h'(x))^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+(0.15x^{0.5})^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+0.0225x}dx$

Let $1+0.0225x=t$

$dx=\frac{1}{0.0225}dt$

$s=\frac{2}{0.0225}\int_{0}^{17.5}\sqrt{t}dt$

$s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}$

$s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}$

$s=\frac{4}{3\times 0.0225}((1+0.0225(17.5))^{\frac{3}{2}-1)$

s=28.21

Weight of cable= $28.21\times 10.4=293.38$ pound

8 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

4x+2y=6 please solve for y

harkovskaia [24]

Answer:

y=3-2x

Step-by-step explanation:

Move all terms that don't contain y to the right and solve.

4 0

3 years ago

What is the algebra expression for this six feet less than the width

Tatiana [17]

So,

"six feet less than"

- 6

"the width"

w

w - 6

"six feet less than the width (w)"

w - 6

4 0

4 years ago

What are the measures of angle 1 and angle 2

aleksley [76]

Answer:

Please attach the image first

8 0

2 years ago

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