28in.=_____ ft ‍ ‍ ‍ ‍ ‍ ‍

What is the length of AC¯¯¯¯¯ ? Enter your answer in the box.

klasskru [66]

In the question, the angles proves that this triangle is an isoceles triangle.

So we can set the two sides equal: $x + 4 = 3x - 8$

Add $-x+8$ to both sides of the equation: $12 = 2x$

Divide both sides by 2: $x = 6$

We find that $x=6$ . AC is also x, so, it is 6 too.

5 0

4 years ago

four sisters bought a present for their mother. They received a 10% discount on the original price of the gift. after the discou

artcher [175]

So first you will need to multiply 9 by 4. You get 36. Now you need to add the 10 percent and you get 39.6! That's the answer!

7 0

4 years ago

Read 2 more answers

What is the answer? I am so lost- thanks for any help given.

aev [14]

Answer:

i think that the answer is 13 meters

Step-by-step explanation:

7 0

3 years ago

Please help, I will mark brainliest if correct.

nadezda [96]

Answer:

100

Step-by-step explanation:

something about them being diagnol to each other makes them equal

havent done geometry in a while lol

hope this helps <3

5 0

3 years ago

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Based on an indication that mean daily car rental rates may be higher for Boston than for Dallas, a survey of eight car rental c

Elina [12.6K]

Answer:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

Step-by-step explanation:

Data given

$n_1 =8$ represent the sample size for group Boston

$n_2 =9$ represent the sample size for group Dallas

$\bar X_1 =47$ represent the sample mean for the group Boston

$\bar X_2 =44$ represent the sample mean for the group Dallas

$s_1=3$ represent the sample standard deviation for group Boston

$s_2=3$ represent the sample standard deviation for group Dallas

We can assume that we have independent samples from two normal distributions with equal variances and that is:

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

Let the subindex 1 for Boston and 2 for Dallas we want to check the following hypothesis:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

The statistic is given by this formula:

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

Where t follows a t student distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

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

Replacing we got:

$\S^2_p =\frac{(8-1)(3)^2 +(9 -1)(3)^2}{8 +9 -2}=9$

And the deviation would be just the square root of the variance:

$S_p=3$

The statitsic would be:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

5 0

4 years ago