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3 years ago

Latrell bought 4 bags of powdered sugar. He got a total of 5 1/2 cups of sugar. How many cups of sugar were in each bag?

1 answer:

5 0

So all 4 bags of sugar yielded 5 and 1/2 cups..... well.. how many in each one? well, let's split the 5 and 1/2 evenly in 4 pieces.

$\bf \stackrel{mixed}{5}\frac{1}{2}\implies \cfrac{5\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{11}{2}}\\\\
-------------------------------\\\\
\cfrac{\quad\frac{11}{2} \quad }{4}\implies \cfrac{\quad\frac{11}{2} \quad }{\frac{4}{1}}\implies \cfrac{11}{2}\cdot \cfrac{1}{4}\implies \cfrac{11}{8}\implies \stackrel{cups}{1\frac{3}{8}
}$

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The chamber of commerce of a Florida Gulf Coast community advertises that area residential property is available at a mean cost

Alja [10]

Answer:

a) In the step-by-step explanation.

b) The null and alternative hypothesis are:

$H_0: \mu=125000\\\\H_a:\mu> 125000$

c) P-value = 0.015

d) The conclusion, at a level of significance of 0.05, is that there is enough evidence to support the claim that that area residential property is available at a mean cost significantly higher than $125,000 per lot.

e) The claim of the chamber of commerce is false. We have statistical evidence against that claim and we can conclude that the mean prices are significantly higher than $125,000.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim, expressed in the alternative hypothesis, is that area residential property is available at a mean cost significantly higher than $125,000 per lot.

Then, the null and alternative hypothesis are:

$H_0: \mu=125000\\\\H_a:\mu> 125000$

The significance level is 0.05.

The sample has a size n=32.

The sample mean is M=130000.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12500.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12500}{\sqrt{32}}=2209.71$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{130000-125000}{2209.71}=\dfrac{5000}{2209.71}=2.26$

The degrees of freedom for this sample size are:

$df=n-1=32-1=31$

This test is a right-tailed test, with 31 degrees of freedom and t=2.26, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t>2.26)=0.015$

As the P-value (0.015) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that that area residential property is available at a mean cost significantly higher than $125,000 per lot.

6 0

3 years ago

A fair coin was flipped 50 times. It landed on heads 29 times.

faust18 [17]

Answer:

21/50, 42%

Step-by-step explanation:

The total times the coin was flipped is 50 and the number of times it landed on tails is 50-29=21. So we need to divide 21 by 50, which gives us 42%.

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3 years ago

At what rate is a car traveling, if it goes 157.5 miles in 2.5 hours?

Finger [1]

divide total miles by time:

157.5 / 2.5 = 63 miles per hour

4 0

3 years ago

PLEASE HELP WITH THIS ONE QUESTION

Nana76 [90]

Answer:

I'm not sure about the factors, but it's reflected over the x axis if they helps narrow it down

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2 years ago

The perimeter of a rectangular baking sheet is 58 inches and its area is 201.25 in.2. What are the length and width of the bakin

anzhelika [568]

Answer:

$Length=17.5\ in\\\\Width=11.5\ in$

Step-by-step explanation:

The formula for calculate the Area of a rectangle is:

$A=lw$

Where "l" is the lenght and "w" is the width.

And the formula for calculate the peimeter of a rectangle is:

$P=2l+2w$

Where "l" is the lenght and "w" is the width.

We know that the perimeter of the rectangular baking sheet is 58 inches and its area is 201.25 in². Then:

$A=201.25\ in^2\\\\P=58\ in$

<u>The steps are:</u>

1. Solve for the "l" from the formula $A=lw$ :

$A=lw\\\\l=\frac{A}{w}\\\\\l=\frac{201.25}{w}$

2. Substitute $l=\frac{201.25}{w}$ into the formula $P=2l+2w$ and solve for "w":

$58=2(\frac{201.25}{w})+2w\\\\58=\frac{402.5}{w}+2w\\\\58=\frac{402.5+2w^2}{w}\\\\58w=402.5+2w^2\\\\2w^2-58w+402.5=0$

Applying the Quadratic formula $x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$ , we get:

$x=w=\frac{-(-58)\±\sqrt{(-58)^2-4(2)(402.5)}}{2(2)}\\\\w_1=17.5\\\\w_2=11.5$

3. Substitute $w_1=17.5$ into $l=\frac{201.25}{w}$ :

$l=\frac{201.25}{17.5}=11.5$

4. Substitute $w_2=11.5$ into $l=\frac{201.25}{w}$ :

$l=\frac{201.25}{11.5}=17.5$

Therefore, since the value of the lenght of a rectangle must be greater that the value of the width, we can conclude that the lenght and the width of the rectangular baking sheet are:

$l=17.5\ in\\\\w=11.5\ in$

8 0

3 years ago