What is the solution set for

A man is 32 years old and his son is 5 years old. How many years hence will the

ArbitrLikvidat [17]

Answer:

Step-by-step explanation:

the current ages:

father= 32 yrs old

son=5 yrs old

-------

4(5+x) = 32+x

20+4x = 32+x

4x-x = 32–20

3x =12

x = 12/3

x = 4 years

------------------------------------

after the 4 years later the farther age will be 36 ( 32+4 = 36)

the son will be 9 ( 5+4= 9 years)

the x presents how the father will be after 4 times the age of the son

3 0

3 years ago

Is anyone good at statistics

masya89 [10]

Im taking stats right now i might be able to help.

4 0

3 years ago

Tamia's family traveled 5/8 of the distance to her grandfather's house on Saturday. They traveled 1/3 of the remaining distance

antoniya [11.8K]

Answer:

2/8

Step-by-step explanation:

Tamila traveled 5/8 of the distance to her grandmother house on Saturday

They traveled 1/3 of the remaining distance on Sunday

Therefore the total distance traveled to her grandmother's house on Sunday can be calculated as follows

5+1

= 6

8-6

= 2

2/8

Hence the distance travelled is 2/8

8 0

2 years ago

1. The box plot shows the ages of people attending a music concert. (4 points)

zhannawk [14.2K]

did you get the answers

8 0

3 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

3 years ago