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3 years ago

How many square yards of carpet needed. to carpet a room that measures 16ft.×22ft

Mathematics

1 answer:

Levart [38]3 years ago

6 0

Assuming by the measurements of the room that is it of rectangular shape.

Rectangle area formula = W ⋅ L

W = 16 ft and L = 22 ft

Area in foot = 16 ⋅ 22 = 352 ft

Converting feet to yard: 352 ft = 117.333 yd

Square yards of carpet needed ≈ 117 $yd^{2}$

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A car rental company charges a one-time application fee of 20 dollars, 45 dollars per day, and 15 cents per mile for its cars. W

crimeas [40]

Answer:

$C(d,m) = 20 + 45d + 0.15m$

Step-by-step explanation:

In this problem, we have that the cost of renting a car C is a function of the number of days d, and the number of miles driven, m.

There is also a fixed cost F.

So the equation for the cost of renting a car is:

$C(d,m) = F + a*d + b*m$

In which a is the daily cost and b is the cost per mile.

A car rental company charges a one-time application fee of 20 dollars, 45 dollars per day, and 15 cents per mile for its cars.

This means that $F = 20, a = 45, b = 0.15$

Write a formula for the cost, C, of renting a car as a function of the number of days, d, and the number of miles driven, m.

$C(d,m) = 20 + 45d + 0.15m$

6 0

3 years ago

A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportio

Reptile [31]

Answer:

The probability is $P(|p-\^{p}| > 0.03) = 0.0040$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.07$

The mean of the sampling distribution is $\mu_p = 0.07$

The sample size is n = 600

Generally the standard deviation is mathematically represented as

$\sigma_p = \sqrt{\frac{p (1 -p)}{n} }$

=> $\sigma_p = \sqrt{\frac{0.07(1 -0.07)}{600} }$

=> $\sigma_p = 0.010416$

Generally the probability that the proportion of airborne viruses in a sample of 600 viruses would differ from the population proportion by greater than 3% is mathematically represented as

$P(|p-\^{p}| > 0.03) = 1 - P(|p -\^{p}| \le 0.03)$

=> $P(|p-\^{p}| > 0.03) = 1 - P(-0.03 \le p -\^{p} \le 0.03 )$

Now add p to both side of the inequality

=> $P(|p-\^{p}| > 0.03) = 1 - P( 0.07-0.03 \le \^{p} \le 0.03+ 0.07 )$

=> $P(|p-\^{p}| > 0.03) = 1 - P(0.04 \le \^{p} \le 0.10 )$

Now converting the probabilities to their respective standardized score

=> $P(|p-\^{p}| > 0.03) = 1 - P(\frac{0.04 - 0.07}{0.010416} \le Z \le \frac{0.10 -0.07}{0.010416} )$

=> $P(|p-\^{p}| > 0.03) = 1 - P(-2.88 \le Z \le 2.88 )$

=> $P(|p-\^{p}| > 0.03) = 1 - [P(Z \le 2.88) - P(Z \le -2.88)]$

From the z-table

$P(Z \le 2.88) = 0.9980$

and

$P(Z \le -2.88) = 0.0020$

$P(|p-\^{p}| > 0.03) = 1 - [0.9980 - 0.0020]$

=> $P(|p-\^{p}| > 0.03) = 0.0040$

7 0

3 years ago

for a pair of triangular earrings, how can you tell if they are similar? how can you tell if they are congruent?

nikitadnepr [17]

In order to be similar, they need to have the same angle measurements, but not the same side measurements. They need to be equiangular, but not equilateral. The lengths if their sides do need to be proportionate, though!

If their sides and angles are the same, they're congruent. In order to be congruent, they need to be exactly the same size and shape. They need to be equilateral AND equiangular.

This question was vague, but I hope this was the answer you were looking for. ❤️

7 0

3 years ago

1.25 = А 103 В 122 55 D 52

DiKsa [7]

Answer:

52 that is my anwer next question please

3 0

3 years ago

In a class of 30 students (x+10) study algebra, (10x+3) study statistics, 4 study both algebra and statistics. 2x study only alg

Vladimir [108]

Answer:

1. The Venn diagrams are attached

2. When the statistics students number = 10·x + 3, we have;

The number of students that study

a. Algebra = 128/11

b. Statistic = 213/11

When the statistics students number = 2·x + 3, we have;

The number of students that study

a. Algebra = 16

b. Statistic = 15

Step-by-step explanation:

The parameters given are;

Total number of students = 30

Number of students that study algebra n(A) = x + 10

Number of students that study statistics n(B) = 10·x + 3

Number of student that study both algebra and statistics n(A∩B) = 4

Number of student that study only algebra n(A\B) = 2·x

Number of students that study neither algebra or statistics n(A∪B)' = 3

Therefore;

The number of students that study either algebra or statistics = n(A∪B)

From set theory we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 10·x + 3 - 4 = 27

11·x+13 = 27 + 4 = 31

11·x = 18

x = 18/11

The number of students that study

a. Algebra

n(A) = 18/11 + 10 = 128/11

b. Statistic

n(B) = 213/11

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 128/11 - 4 = 84/11

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 213/11 - 4 = 169/11

However, assuming n(B) = (2·x + 3), we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 2·x + 3 - 4 = 27

2·x+3 + x + 10= 27 + 4 = 31

3·x = 18

x = 6

Therefore, the number of students that study

a. Algebra

n(A) = 16

b. Statistics

n(B) = 15

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 16 - 4 = 12

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 15 - 4 = 11

The Venn diagrams can be presented as follows;

6 0

3 years ago