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hey the question is, " The length of a rectangle is twice its width. The perimeter of the rectangle is 126 feet."

Tema [17]

Answer:

42 = l

21 = w

Step-by-step explanation:

{l = 2w

{126 = 2w + 2l

126 = 2w + 2[2w]

126 = 2w + 4w

126 = 6w

21 = w [Plug this back into both equations to get the length of 42]; 42 = l

I am joyous to assist you anytime.

4 0

3 years ago

Please help me!! I will mark brainliest

Annette [7]

I drew it out below (don't you just love my effort? XD)

You're gonna use the sine trig function to find the length of the shadow

sin26 / 1 = 43 / x; now you cross multiply

sin26 x = 43; divide both sides by sin26

x = 43 / sin26

x = 98 which is the shadow

Hope this helps!

6 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

A certain high school has 1,488 students. If you subtract the number of girls from boys you get 78. Let x be the number of boys

Dahasolnce [82]

Answer:

boys = 705

girls = 783

Step-by-step explanation:

We are to find the number of boys and girls in the school

the answer can be determined using simultaneous equation

x = boys

y = girls

y - x = 78 equation 1

y + x = 1488 equation 2

subtract equation 1 from 2

2x = 1410

x = 1410/2 = 705

Substitute for x in equation 1

y - 705 = 78

y = 705 + 78 = 783

3 0

2 years ago

What number is 54% of 125 type an integer or a decimal round to the nearest tenth

eimsori [14]

Answer is 67.5

To get this you must first turn 54% into a decimal.
54%---> .54
125*.54= 67.5
So 54% of 125 is 67.5

4 0

3 years ago

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