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4 years ago

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Clare is painting some doors that are all the same size. She used 2 liters of paint to cover 1 3/5 doors. How many liters of pai

nt are needed or 1 door?

1 answer:

barxatty [35]4 years ago

8 0

Answer:

$1\frac{1}{4}\ liters$

Step-by-step explanation:

step 1

Convert mixed number to an improper fraction

$1\frac{3}{5}=\frac{1*5+3}{5}=\frac{8}{5}$

step 2

Using proportion

$\frac{2}{(8/5)}\frac{liters}{doors} =\frac{x}{1}\frac{liters}{door}\\ \\x=2/(8/5)\\ \\x=10/8\ liters$

step 3

Convert to mixed number

$\frac{10}{8}\ liters=\frac{8}{8}+\frac{2}{8}=1\frac{1}{4}\ liters$

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Help help help help help

dolphi86 [110]

Answer:

x = <u>12/5</u>

Step-by-step explanation:

Since the original is 10 and the second is 6, that means that the scale is 10:6 or 5:3. It can also be related as %60 of the length.

<u>Step 1: Make an equation</u>

10/6 = 4/x

<u>Step 2: Cross Multiply</u>

10x = 24

<u>Step 3: Divide</u>

10x / 10 = 24 / 10

24/10 = 12/5

x = 2.4

5 0

3 years ago

Read 2 more answers

henry and his brother each start a savings account. henry begins with $200 and deposits $25 each month. his brother begins with

zlopas [31]

Answer:

Henry after 2 months: 250

His brother after 2 months: 220

Step-by-step explanation:

Henry started with $200 so add $25 twice (for 2 months) and get $250 dollars after 2 months.

His brother started with $150 so add $35 twice and get $220 dollars after 2 months

4 0

3 years ago

A van,a moving company charges 50.00 plus 5.50 by per mile. The table shows the total cost for the number of miles drivin

Anna71 [15]

y = 55/10x + 50 is the equation to represent this situation

8 0

2 years ago

Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his custom

ira [324]

Answer:

a) The best estimator for the population mean is given by the sample mean $\hat \mu = \bar X = 5.70$

b) The 99% confidence interval would be given by (4.969;6.431)

c) We are 99% confident that the true mean for th number of gallons of gasoline sold to his customers is between 4.969 and 6.431.

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=1.9$ represent the population standard deviation

n=45 represent the sample size

2) Part a

The best estimator for the population mean is given by the sample mean $\hat \mu = \bar X = 5.70$

3) Part b. Develop a 99 percent confidence interval for the population mean.

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$5.7-2.58\frac{1.9}{\sqrt{45}}=4.969$

$5.7+2.58\frac{1.9}{\sqrt{45}}=6.431$

So on this case the 99% confidence interval would be given by (4.969;6.431)

4) Part c. Interpret the meaning of part (b).

On this case we can say this: We are 99% confident that the true mean for th number of gallons of gasoline sold to his customers is between 4.969 and 6.431.

4 0

4 years ago

PLEASE HELP ASAP -- WITH THE WORKK<br><br>50 points

MAXImum [283]

Part-A:-

<h3>(f×g)=</h3>

$\\ \sf\longmapsto x^2+3x-70(4x-12)$

$\\ \sf\longmapsto 4x^2+12x-280-12x^2+36x+840$

$\\ \sf\longmapsto -8x^2+48x+560$

Now

$\\ \sf\longmapsto (f.g)(3)$

$\\ \sf\longmapsto -8(3)^2+48(3)+560$

$\\ \sf\longmapsto -72+144+560$

$\\ \sf\longmapsto 72+560$

$\\ \sf\longmapsto 632$

Part-B:-

$\\ \sf\longmapsto f(3)$

$\\ \sf\longmapsto (3)^2+3(3)-70$

$\\ \sf\longmapsto 9+9-70$

$\\ \sf\longmapsto 18-70$

$\\ \sf\longmapsto -52$

Now

$\\ \sf\longmapsto g(f(3))$

$\\ \sf\longmapsto g(-52)$

$\\ \sf\longmapsto 4(-52-3)$

$\\ \sf\longmapsto 4(-55)$

$\\ \sf\longmapsto -220$

Part:-C

$\\ \sf\longmapsto g(-2)$

$\\ \sf\longmapsto 4(-2-3)$

$\\ \sf\longmapsto 4(-5)$

$\\ \sf\longmapsto -20$

Now

$\\ \sf\longmapsto f(g(-2))$

$\\ \sf\longmapsto f(-20)$

$\\ \sf\longmapsto (-20)^2+3(-20)-70$

$\\ \sf\longmapsto 400-60-70$

$\\ \sf\longmapsto 400-130$

$\\ \sf\longmapsto 270$

Part-D:-

g×f=f×g

$\\ \sf\longmapsto (g.f)(-2)$

$\\ \sf\longmapsto -8(-2)^2+48(-2)+560$

$\\ \sf\longmapsto 32-96+560$

$\\ \sf\longmapsto -64+560$

$\\ \sf\longmapsto 496$

6 0

3 years ago