F(x) = 5x - 3<br><br> F(5)= ????

yaroslaw [1]

They are = because both can be reduced to 9/17
Also their cross products are equal 27 x 34 = 18 x 51 - Both equal 918 so both fractions are equal to each other.

8 0

3 years ago

Read 2 more answers

Find the surface area of each prism

goldfiish [28.3K]

Answer:

Step-by-step explanation:

2.

Surface area = ( 6 *10 ) *3 + ( ( 7*4.87) /2)*2 = 180 + 34.09 = 214.09

3.

Surface area = (5*10) *3 + (( 7*4.2)/2)*2 = 150 +29.4 = 179.4

6 0

3 years ago

solve the average time for a student to complete this quiz is 24 24 minutes with a standard deviation of 2 minutes. What is the

LuckyWell [14K]

Answer:

The correct answer is "(18, 30) minutes".

Step-by-step explanation:

Mean,

= 24

Standard deviation,

= 2

The range of minutes where 99.7% of individuals completed the quiz will be:

= $Mean \pm 3\times Stdev$

On substituting the values, we get

= $24 \pm 3\times 2$

= $24 \pm 6$

now,

= $24-6$

= $18$

Or,

= $24+6$

= $30$

i.e., 18 to 30 minutes

3 0

3 years ago

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

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

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago

Please help. thanks.

Free_Kalibri [48]

Hello!

You have to find the area of the prism and the pyramid

rectangular prism is lwh

Put in the values you know

15 * 5 * 7 = 525

For a prism you use

$\frac{lwh}{3}$

Put in the values you know

$\frac{15 * 5 *13 }{3}$

Multiply

975/3

Divide

325

Add the values you know

325 + 525 = 850

The answer is D) $850cm^{3}$

Hope this helps!

5 0

3 years ago

F(x) = 5x - 3 F(5)= ????