Evaluate the following expression:<br> [9 (10

Someone give me the answer plssssss

salantis [7]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

2ft 3in = how many inches

Arte-miy333 [17]

1 ft = 12 inches

2 ft = 24 inches

24 inches + 3 inches = 27 inches

7 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

The length of a rectangle is 12 units longer than the width. The perimeter is 7 times the width. Find the length and the width o

katen-ka-za [31]

Log explanation below; answer is at bottom.

If the length, l, is 12 units longer than the width, w, then w = l - 12. If the perimeter, p,

p = 2l +2w,

is 7 x w, then

w = p/7

When two things are set equal to the same variable, they are equal to each other, so,

l - 12 = p/7

Now you need to get rid of the p so you are only working with one variable. To do this you plug in whatever p is equal to for p, so,

l - 12 = (2l + 2w)/7 now to get rid of the w do the same thing we did with p just for w. So,

l - 12 = (2l + 2(l - 12))/7

To solve this you want to multiply both sides by 7 first to get rid of the fraction.

7l - 84 = 2l + 2(l - 12)

Next you want to distribute the 2 over the l and the 12.

7l - 84 = 2l + 2l - 24

Next you want to combine like terms on each side.

7l - 84 = 4l - 24

Next add 84 and subtract 4l from sides to isolate the variable.

3l = 60

Now divide each side by 11 to get your answer.

l = 20.

To find the width,

w = l - 12

Just plug in and solve.

w = 20 - 12

w = 8

So your length and width are 20 and 8.

l = 20

w = 8

7 0

4 years ago

Someone please help ?

ikadub [295]

a. 1490

5 x 37.25= 186.25

186.25 x 8 = 1,490

8 0

4 years ago

Evaluate the following expression: [9 (10 – 2)] = [2 + (5 x 2)]