Ask question

Ask question

All categories

bagirrra123 [75]

2 years ago

3

Lucy decides to paint only the walls of 3 bedrooms in her house. Each room has a door, a closet door, and a window. She is only

applying 1 coat of paint to the walls. Each room is 14 feet by 12 feet with 8-foot ceilings. Each door has dimensions 6.5 feet by 2.5 feet. Each window has dimensions 4.5 feet by 3 feet. 1 gallon of paint with cover 350 square feet. Paint is sold in 1-gallon containers. Enter the minimum number of 1-gallon cans of paint she need.

1 answer:

Otrada [13]2 years ago

8 0

The minimum number of 1-gallon cans of paint Lucy needs is 5.

Step-by-step explanation:

Given in the question,

There are 3 bedrooms of dimensions (14*12*8)feets

Thus total area of the room = area of walls+area of 1 ceiling

= (2*area of opposite walls)+(2*area of another set of opposite walls)+ceiling

=(2*14*8)+(2*12*8)+(14*12)

=224+192+168

=584 square feet.

Now deducting the area of doors and windows of area (6.5*2.5)sq ft and (4.5*3)sq ft;

=584-(6.5*2.5)-(4.5*3)

=584-16.25-13.5

=554.25 sq ft.

There are 3 such of bedrooms.Thus;

=3*554.25

=1662.75 sq ft.

Now dividing it with 350 sq ft which is the capaciy paint for 1-gallon container;

=1662.75/350

=4.7507.

which is nearly equals to 5. Thus,the minimum number of 1-gallon cans of paint Lucy needs is 5.

You might be interested in

LORAN is a long range hyperbolic navigation system. Suppose two LORAN transmitters are located at the coordinates (-100,0) and (

Gre4nikov [31]

Answer:

$\frac {(x)^{2}}{8100}+\frac {(y)^{2}}{1900}=1$

Step-by-step explanation:

center of the hyperbola is (0,0) = (h, k)

c = the distance form the center to either focal point = 100

$c^{2} =100^{2}=10000$

The differences from the receiver to the transmitters = 2a

2a = 180 miles

a = 180/2=90 miles

$a^2 =90^{2}= 8100$

$b^{2}= c^{2} - a^{2}$

$b^{2}= 100^{2}-90^{2}=1900$

$b^{2}=1900$

The standard form is

$\frac {(x-h)^{2}}{a^{2}}+\frac {(y-k)^{2}}{b^{2}}=1$

$\frac {(x-0)^{2}}{a^{2}}+\frac {(y-0)^{2}}{b^{2}}=1$

$\frac {(x)^{2}}{8100}+\frac {(y)^{2}}{1900}=1$

3 0

2 years ago

Candidate A is facing two opposing candidates in a mayoral election. In a recent poll of 300 residents, 98 supported candidate B

babymother [125]

Answer:

[0.4235, 0.5365]

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

X=300-98-58=144 represent the people that support the candidate A in the sample

$\hat p=\frac{144}{300}=0.48$ estimated proportion of people that support the candidate A in the sample

$\alpha=0.05$ represent the significance level

Confidence =0.95 or 95%

p= population proportion of people that support the candidate A.

Confidence 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:

$0.48 - 1.96 \sqrt{\frac{0.48(1-0.48)}{300}}=0.4235$

$0.48 + 1.96 \sqrt{\frac{0.48(1-0.48)}{300}}=0.5365$

And the 95% confidence interval would be given (0.4235;0.5365).

$0.4235 \leq p \leq 0.5365$

8 0

2 years ago

There are 451 children in lias write number greater than 451

Fed [463]

452 ?

i think your question doesnt make much sense

3 0

2 years ago

The cost of producing x soccer balls in thousands of dollars is represented by h(x) = 5x + 6. The revenue is represented by

Firdavs [7]

Answer:

(k-h)(x) = 4x - 8

Step-by-step explanation:

We know Profit = Revenue - Cost

Basically we gotta subtract cost function from revenue function and get the profit function.

The cost function is h(x) = 5x + 6

The revenue function is k(x) = 9x - 2

Hence, Profit is:

(k-h)(x) = (9x - 2) - (5x + 6)

(k-h)(x) = 9x -2 - 5x - 6

(k-h)(x) = 4x - 8

3 0

2 years ago

Read 2 more answers

Traffic speed: The mean speed for a sample of cars at a certain intersection was kilometers per hour with a standard deviation o

aliina [53]

Answer:

Step-by-step explanation:

Hello!

X₁: speed of a motorcycle at a certain intersection.

n₁= 135

X[bar]₁= 33.99 km/h

S₁= 4.02 km/h

X₂: speed of a car at a certain intersection.

n₂= 42 cars

X[bar]₂= 26.56 km/h

S₂= 2.45 km/h

Assuming

X₁~N(μ₁; σ₁²)

X₂~N(μ₂; σ₂²)

and σ₁² = σ₂²

<em>A 90% confidence interval for the difference between the mean speeds, in kilometers per hour, of motorcycles and cars at this intersection is ________.</em>

The parameter of interest is μ₁-μ₂

(X[bar]₁-X[bar]₂)± $t_{n_1+n_2-2}$ * $Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$

$t_{n_1+n_2-2;1-\alpha /2}= t_{175; 0.95}= 1.654$

$Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{134*16.1604+41*6.0025}{135+42-2} } = 3.71$

[(33.99-26.56) ± 1.654 *( $3.71*\sqrt{\frac{1}{135} +\frac{1}{42} }$ )]

[6.345; 8.514]= [6.35; 8.51]km/h

<em>Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15</em>

$t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485$

$Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06$

[(475.12-321.34) ± 2.485 *( $33.06*\sqrt{\frac{1}{12} +\frac{1}{15} }$ )]

[121.96; 185.60]

I hope this helps!

3 0

2 years ago