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3 years ago

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The five members of the Traynor family each buy train tickets. During the train ride, each family member buys a boxed lunch for

$6.50. If the total cost of the trip is $248.50, what is the price of each train ticket?

1 answer:

dmitriy555 [2]3 years ago

8 0

Answer:

Step-by-step explanation:

it’s B

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I have over 20 pencils in my desk

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Which equation represents the line that is perpendicular to graph of 4x+3y=9 and passes through (-2,3)

goldfiish [28.3K]

4y = 3x + 18

Step-by-step explanation:

NOTE THAT A line that is perpendicular to another has a negative inverse of the slope of the other line. The products of their slopes, that is, is always -1

Therefore we can begin by finding the slope of this line defined by the function 4x+3y=9

3y = -4x + 9

y = -4/3 x + 9/3

y = -4/3 x + 3

The slope of the perpendicular line is, therefore;

¾ - this is the negative inverse of -4/3

Now that we know the slope, we need to find the y-intercept. This is where x = 0 and the line meets the y-axis;

i.e (0, y)

The other given point, where the line crosses is (-2, 3). Remember that to get the gradient we use the formula;

Gradient = Δ y / Δ x

¾ = (3 – y) / (-2 – 0)

¾ = (3 –y) / -2

¾ * -2 = 3 – y

-3/2 = 3 – y

-3/2 – 3 = -y

9/2 = y ←– This is the y-intercept

Remember the function of a straight line is given;

y = mx + c (m being slope and c being y-intercept)

y = 3/4 x + 9/2

4y = 3x + 18

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You work in the HR department at a large franchise. you want to test whether you have set your employee monthly allowances corre

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Answer:

1) Null hypothesis: $\mu \leq 500$

Alternative hypothesis: $\mu > 500$

$z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90$

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

$z_{crit}= 2.33$

For this case we see that the calculated value is higher than the critical value

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

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

$p_v =P(z>5.90)=1.82x10^{-9}$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

Step-by-step explanation:

Part 1

Data given

$\bar X=640$ represent the sample mean

$\sigma=150$ represent the population standard deviation

$n=40$ sample size

$\mu_o =500$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Step1:State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is higher than 500, the system of hypothesis would be:

Null hypothesis: $\mu \leq 500$

Alternative hypothesis: $\mu > 500$

Step 2: Calculate the statistic

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

We can replace in formula (1) the info given like this:

$z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90$

Step 3: Calculate the critical value

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

$z_{crit}= 2.33$

Step 4: Compare the statistic with the critical value

For this case we see that the calculated value is higher than the critical value

Step 5: Decision

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

Part 2

P-value

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

$p_v =P(z>5.90)=1.82x10^{-9}$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

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