Write the slope-intercept form of the equation of each line. identify the slopes y-intercept.

The CEO of the Jen and Benny's ice cream company is concerned about the net weight of ice cream in their 50 ounce ice cream tubs

MissTica

Answer:

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

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

$p_v =P(t_{25}>1.8025)=0.0418$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=54.98$ represent the sample mean

$s=8.43$ represent the sample deviation

$n=26$ sample size

$\mu_o =52$ represent the value that we want to test

$\alpha=0.05$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 52, the system of hypothesis would be:

Null hypothesis: $\mu \leq 52$

Alternative hypothesis: $\mu > 52$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

P-value

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

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

$p_v =P(t_{25}>1.8025)=0.0418$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

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The length of the shorter base in an isosceles trapezoid is 4 in, its altitude is 5 in, and the measure of one of its obtuse ang

siniylev [52]

The sum of angles in any quadrilateral, including trapezoid, is 360⁰.
Because we have <span>an isosceles trapezoid, we have 2 angles with measure 135⁰,
and we have 2 equal acute angles with measure x⁰.
So, we can find value of acute angle,
135*2 +2x =360⁰
270+2x=360
2x=360-270
2x=90
x=45⁰

So, acute angles in trapezoid = 45⁰.
From triangle ABC,
angle ACB =90⁰
angle A=45⁰,
so angle ABC= 180-(90-45)=45⁰
Triangle ABC is isosceles triangle,so |AC| = |CB|= 5 in.

So, longer base AA' = 5+4+5= 14 in

Now, we can find area of trapezoid.
shorter base = 4 in
longer base = 14 in
altitude =h = 5 in
Area of trapezoid =(1/2)(base1+base2)*h
Area of trapezoid = (1/2)(4+14)*5= 9*5=45 in²
Answer is 45 in².</span>