Please help me!!

In the diagram, if WA = 8, AX = 5, and YA = 10, then the length of YZ is

Zinaida [17]

Answer:

4

Step-by-step explanation:

WA times AX = YA x YZ

8x5=40

because they have to equal each other we just have to divide 40 by 10

40/10=4 so segment YZ=4

5 0

2 years ago

Hip hop music makes up 75% of Landon's music library. If he has downloaded 120 hip hop songs, how many songs does Landon have in

Otrada [13]

Answer: 160 songs.

Step-by-step explanation:

75% hip hop music - 120 songs

100% all music - x songs

Let's write the proportion:

$\dfrac{120}{75} =\dfrac{x}{100} \\\\75x=120 \cdot 100\\75x=12000\\75x : 75 = 12000 : 75\\x = 160$

Thus, Landon have 160 songs in his music library.

4 0

3 years ago

10-2(2x+1)=4(x-2)solve each equation for x graph your Solution on a number line

stepladder [879]

Answer:

No solution

Step-by-step explanation:

Actually, you have only one equation here. We will solve it for x.

Performing the indicated multiplication, we obtain:

10-2(2x+1)=4(x-2) => 10 - 4x - 2 = 4x - 8, or 10 - 2 = -8. This can never be true, so this equation has no solution.

6 0

3 years ago

Read 2 more answers

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

A computer code is based on the prime factorization of 160.Find the prime factorization of 160

lilavasa [31]

Ndhdjjdjdjshduhdjskakksjdnfbbfbbnfbbbdbdvdbdbbdbdbfhf

8 0

3 years ago