Which equation matches the graph of the greatest integer function given below?

martin uses 5/8 of a gallon of paint to cover 4/5 of a wall. what is the unit rate at which martin paints in walls per gallon?

sergeinik [125]

So there are 5/8 gallons and 4/5 walls to cover, so the wall/gallon rate will just be 45 ÷ 5/8.

$\bf \cfrac{\quad \frac{4}{5}~wall\quad }{\frac{5}{8}~gallon}\implies \cfrac{4}{{5}}\cdot \cfrac{8}{{5}}\implies \cfrac{32}{25}~\cfrac{wall}{gallon}\implies 1.28~\frac{wall}{gallon}$

7 0

2 years ago

Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in

likoan [24]

Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=7 represent the households with three or more large-screen TVs

$\hat p=\frac{7}{500}=0.014$ estimated proportion of households with three or more large-screen TVs

$\alpha=0.05$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

Part b

Yes is possible. We hav that $np>10$ and $n(1-p)>10$ so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370$

$0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243$

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

4 0

2 years ago

On frank's bus ride to work, all available seats are filled and 6 passengers are standing in the aisles. if 10 passengers get of

avanturin [10]

<span>Answer: 4 Explanation: Initially 6 people are standing. 10 people get off the bus. The six people who are standing will take their seats and remaining 4 seats will be occupied by four out of the eight people who alight the bus. So now 4 people will be standing on the aisles.</span>

5 0

1 year ago

Read 2 more answers

bookstore ordered 1,528 packs of pens and 1,823 packs of pencils at the prices shown. how much did the bookstore spend on pens?

IgorC [24]

I found that the given prices are $ 4 per pack of pens and $ 3 per pack of penciles.

The question is how much the bookstore spent on pens.

Then you have to multiply the number packs of pens, which is 1,528, times the price per pack pens which is $ 4.

So, the answer is: 1528 packs of pens * $ 4 / pack of pens = $6112.

8 0

2 years ago

A company purchased a delivery van for $28,000 with a salvage value of $3,000 on September 1, Year 1. It has an estimated useful

Tems11 [23]

Answer:

Depreciation till December 31, Year 1 will be equal to 1,250$

Step-by-step explanation:

Purchasing Cost = 28,000$

Salvage Value = 3,000$

Total Depreciation:

Total Depreciation over 5 years (60 Months) = Purchasing Cost - Salvage Value

Total Depreciation over 5 years (60 Months) = 28,000 - 3,000

Total Depreciation over 5 years (60 Months) = 25,000$

Monthly Depreciation:

Using the unity method we have monthly depreciation by dividing the total depreciation by the total no. of months as below:

Total Depreciation over a single month =25,000/60

Total Depreciation over a single month = 416.67$ (Monthly Depreciation)

Depreciation till December 31, Year 1

As from September 1, Year 1 to December 31, Year 1, its been 3 months therefore total depreciation will be = 3 * Monthly Depreciation

Depreciation till December 31, Year 1 = 3 * 416.67

Depreciation till December 31, Year 1 = 1,250$

4 0

2 years ago

Which equation matches the graph of the greatest integer function given below?​

Which equation matches the graph of the greatest integer function given below?