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kherson [118]
3 years ago
12

While painting the interior of a large house, Gene is going to transfer some paint in a large 5 gallon container to a smaller bu

cket. Unfortunately he slips and spills 2 gallons of paint all over the floor. If each wall in the house takes 9/24 of a gallon to paint, how many walls can he paint with what he has left?
Mathematics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

8 walls

Step-by-step explanation:

While painting the interior of a large house, Gene is going to transfer some paint in a large 5 gallon container to a smaller bucket. Unfortunately he slips and spills 2 gallons of paint all over the floor.

The remaining amount of paint is

= 5 gallons - 2 gallons

= 3 gallons

If each wall in the house takes 9/24 of a gallon to paint, how many walls can he paint with what he has left?

Hence:

9/24 gallon = 1 Wall

3 gallons = x wall

Cross Multiply

9/24 × x = 3 × 1

x= 3 ÷ 9/24

x = 3 × 24/9

x = 8 walls

He can paint 8 walls with what is left.

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29.8 would be the number, but rounded would be 30
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Simplify (6g4h5)2<br> What is the answer
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"A rectangle has a height of 5 cm and its base is increasing at a rate" of 3/2 cm/min. When its area is 60 cm2, at what rate is
sukhopar [10]

Answer:

The diagonal is increasing at the rate of 119/104cm/min of the given rectangle.

Step-by-step explanation:

Dimensions of the rectangle

Height = 5cm

Rate of base = 3/2 cm/min

Area = 60cm^2

We know the area of a rectangle of given by = base* Height

b*h = 60

b*5 = 60

b = 12cm

Applying Pythagoras theorem while drawing a diagonal to the rectangle

  b^2 +h^2 =  D^2\\

 5^2 +12^2 = 13^2

so our diagonal will be 13cm  

Upon differentiating the area of the rectangle  we get

b*h = A=60cm^2

using  the chain rule of differentiation

h*db/dt + b*dh/dt  = 0

b*dh/dt = -h*db/dt

12*dh/dt = -5*3/2

dh/dt = -5/8 cm//min

so the height of the rectangle is decreasing at the rate of -5/8cm/min

now we have all the measurements we need

b = 12 , db/dt = 3/2cm/min

h = 5 , dh/dt = -5/8 cm/min

b^2 +h^2  = D^2

Upon differentiating we get

2b*db/dt + 2h*dh/dt = 2D*dD/dt

b*db/dt + h*dh/dt = D*dD/dt

12*3/2 + 5*(-5/8) = 13*dD/dt

18 -25/8 = 13*dD/dt

\frac{144-25}{8} = 13*dD/dt

dD/dt = \frac{119}{104} cm/min

Therefore the diagonal is increasing at the rate of 119/104cm/min of the given rectangle.

6 0
3 years ago
ccording to the U.S. National Weather Service, at any given moment of any day, approximately 1000 thunderstorms are occurring wo
viva [34]

Answer:

The critical value of <em>z</em> for 99% confidence interval is 2.5760.

The 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

Step-by-step explanation:

Let <em>X</em> = number of lightning strikes on each day.

A random sample of <em>n</em> = 23 days is selected to observe the number of lightning strikes on each day.

The random variable <em>X</em> has a sample mean of, \bar x=8.1\ mn and the population standard deviation, \sigma=0.51\ mn.

The (1 - <em>α</em>)% confidence interval for population mean <em>μ</em> is:

CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

Compute the critical value of <em>z</em> for 99% confidence interval is:

z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.5760

*Use a <em>z</em>-table.

The critical value of <em>z</em> for 99% confidence interval is 2.5760.

Compute the 99% confidence interval for population mean number of lightning strike as follows:

CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\=8.1\pm2.5760\times\frac{0.51}{\sqrt{23}}\\=8.1\pm0.2738\\=(7.8262, 8.3738)\\\approx(7.83, 8.37)

Thus, the 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

The 99% confidence interval for population mean number of lightning strike  implies that the true mean number of lightning strikes lies in the interval (7.83 mn, 8.37 mn) with 0.99 probability.

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3 years ago
Given the below sequence:-1, -3, -5, -7, . . .
vekshin1

Answer:

A. The next 3 terms are —9, —11, —13

B. The sequence is Arithmetic

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Common difference = T2 — T1 or = T3 — T2

=T2 — T1 = —3 —(—1) = —2 or

= T3 — T2 = —5 —(—3) = —2

D. Tn = a + d(n—1)

a = first term = —1

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T27 = —1 + —2(27—1)

T27 = —1 + —2(26)

T27 = —1 —52

T27 = —53

3 0
3 years ago
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