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2 years ago

the blue line and the yellow line trains just arrived at the station. When will they next arrive at the station at the same time

Mathematics

1 answer:

ivann1987 [24]2 years ago

5 0

Could you add some more details about the problem?

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Can you guys help me on question 8 and 9 :)

Reika [66]

Answer:

for 8 i think its a b and c sorry if am wrong

8 0

2 years ago

Devin has a snake. The snake eats one mouse every week which Devin purchases from the local pet store. If m represents the cost

n200080 [17]

Assuming it's not a leap year,
then there are 365 days a year.
This is about 52 weeks.
The cost of a mouse is 52m

4 0

2 years ago

A museum sold 95 tickets for $1,025. Each adult ticket was $15, and each child ticket was $7. Let a represent the number of adul

Genrish500 [490]

Answer:

a + c = 95........eq1

15a + 7c = 1,025....eq2

Step-by-step explanation:

Given:

Total number of ticket sold = 95

Total amount collected = $1,025

Price of each adult ticket = $15

Price of each child ticket = $7

Computation:

Number of adult tickets sold = a

Number of child tickets sold = c

So,

Number of adult tickets sold + Number of child tickets sold = Total number of ticket sold

a + c = 95........eq1

and

15a + 7c = 1,025....eq2

4 0

2 years ago

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who w

dimulka [17.4K]

Answer:

99% CI:

$-4.637\leq\mu_v-\mu_o\leq 3.737$

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}$

The MSE can be calculated as

$MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65$

The weighted sample size nh can be calculated as

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16$

The standard deviation then becomes

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623$

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

$\Delta M=M_v-M_o=5.10-5.55=-0.45$

Then the confidence interval can be constructed as

$\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737$

8 0

2 years ago

Hello there I need help.

andriy [413]

Answer:

A. 40

Step-by-step explanation:

Q1 - 40

Q2 - 65

Q3 - 80

And 90 is max

So 40 is the correct answer.

5 0

2 years ago