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4 years ago

Brainliesttttttttttttt!!!!

Mathematics

2 answers:

max2010maxim [7]4 years ago

8 0

<h2>Answer:</h2>

The slope of the model represent:

The number of cereal boxes sold per day

<h2>Step-by-step explanation:</h2>

The table which represent the days and number of cereal boxes sold is given by:

Days 5 3 8 6 0 4

No. of cereal boxes sold 50 70 20 40 100 60

We know that the slope of a graph is the ratio of change in dependent variable to the change in independent variable.

Here the independent variable is number of days

and the dependent variable is number of cereal boxes sold.

Hence, the slope is:

The number of cereal boxes sold per day.

( since, by the change in number of cereal boxes to the change in number of days we may obtain the per day number of boxes sold)

yawa3891 [41]4 years ago

5 0

It's d)
if you divide the total number of boxes sold by the total amount of days, you get by definition, the number of boxes sold per day

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3 years ago

Read 2 more answers

Solve for P. <br><br> S=A+1/2PL

Dovator [93]

9514 1404 393

Answer:

(b) P = 2(S -A) ÷ L

Step-by-step explanation:

Undo what is done to P.

$S=A+\dfrac{1}{2}PL\qquad\text{given}\\\\S-A = \dfrac{1}{2}PL\qquad\text{subtract $A$}\\\\\dfrac{2}{L}(S-A)=P \qquad\text{multiply by $\dfrac{2}{L}$}\\\\\boxed{P=\dfrac{2(S-A)}{L}}$

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3 years ago

A certain railway company claims that its trains run late 5 minutes on the average. The actual times (minutes) that 10 randomly

iragen [17]

Answer:

Option C) We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 5 minutes

Sample mean, $\bar{x}$ = 9.130 minutes

Sample size, n = 10

Alpha, α = 0.01

Sample standard deviation, s = 1.4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 5\text{ minutes}\\H_A: \mu \neq 5\text{ minutes}$

We have to construct 99% confidence interval.

99% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.01} = \pm 3.249$

$9.130 \pm 3.249(\dfrac{1.4}{\sqrt{10}} ) = 9.130 \pm 1.4383 = (7.6917,10.5683)$

99% Confidence interval: (7.6917,10.5683)

Since the population mean does not lie in he calculated confidence interval, thus, we fail to accept the null hypothesis and accept the alternate hypothesis.

Option C) We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.

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3 years ago