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3 years ago

If one gallon of paint covers 825 sq.ft, how much paint is needed to cover 2640sq.ft. ?

1 answer:

3 0

2640/825=3.2
There is need 4 gallons to cover 2640 sq ft.
80% of the fourth gallon can be used for other purposes after painting.

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Find the percent of change. Original: 80, new: 64

lara31 [8.8K]

Hi there!

To find the percent of change we can use the following formula:
$\frac{new - old}{old} \times 100\%$

Let's fill in our data
$\frac{64 - 80}{80} \times 100\% =$

Subtract
$\frac{ - 16}{80} \times 100\% =$

Divide
$- 0.2 \times 100\%$

And finally multiply
$- 20\%$

Hence, the percent of change is negative 20%.
~ Hope this helps you!

3 0

3 years ago

According to the last census (2010), the mean number of people per household in the United States is LaTeX: \mu = 2.58 Assume a

Veseljchak [2.6K]

Answer:

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, $\mu$ = 2.58 and Standard deviation, $\sigma$ = 0.75

Also, a random sample (n) of 110 households is taken.

Let Xbar = sample mean household size

The z score probability distribution for sample mean is give by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)

P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar $\leq$ 2.50)

P(Xbar < 2.66) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{2.66-2.78}{\frac{0.75}{\sqrt{110} } }$ ) = P(Z < -1.68) = 1 - P(Z 1.68)

= 1 - 0.95352 = 0.04648

P(Xbar $\leq$ 2.50) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{2.50-2.78}{\frac{0.75}{\sqrt{110} } }$ ) = P(Z $\leq$ -3.92) = 1 - P(Z < 3.92)

= 1 - 0.99996 = 0.00004

Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046

7 0

3 years ago

How many units long is AC?

Nonamiya [84]

The answer is B
6 units

3 0

3 years ago

An athlete runs every day for one week. The following week she runs one mile further than she ran on that day the previous week.

cluponka [151]

Answer:

Buy $HBAR

Step-by-step explanation:

You can find it on exchanges like Bittrex, Voyager, Uphold, and soon enough Coinbase!

5 0

3 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0

3 years ago