On an algebra test, the highest grade was 42 points higher than lowest grade.

Identify the property: 3(9) + 3(7) = 3(9 + 7)

Aleks [24]

3(9) = 27
3(7) = 21
3(9+7) = 48
27+21 = 48
48=48

4 0

3 years ago

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What value of x makes the equation true?<br> 5(x-3)-2x=6

svlad2 [7]

5x-15-2x=6
3x=21
x=7

Therefore, x must equal 7.

7 0

2 years ago

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ANSWER THIS CORRECTLY FOR BRAINLIEST!!!!<br><br> This is 8th grade<br> algebra btw

gulaghasi [49]

Answer:

V = 2143.57

Step-by-step explanation:

To find the volume of a sphere you need to use this formula:

$V = \frac{4}{3}\pi r^{3}$

r = radius

Solve:

You need the radius from the diameter. Divide 16/2 which is

8.

$V = \frac{4}{3}(3.14) 8^{3}$

V = 2143.57

7 0

1 year ago

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Even more trig ratios see attachment

egoroff_w [7]

Step-by-step explanation:

1. Use the Pythagorean theorem to solve for side KM:

16^2+KM^2=34^2
256+KM^2=1156
KM^2=900
KM=30

2. Cosine is adjacent over hypotenuse, so cosine of M would be KM/34, or 30/34

3. Tangent is opposite over adjacent, so tangent of L will be KM/16, or 30/16

4. Sine is opposite over hypotenuse, so sine of M will be 16/34

5. KM=30, solved for in step 1.

hope this helps!!

4 0

2 years ago

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