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aleksklad [387]

3 years ago

9

George is building a table out of boards that are 3.75 inches wide. He wants the table to be at least 36 inches wide. How many b

oards does he need to buy? Round to the nearest whole number

1 answer:

balu736 [363]3 years ago

7 0

Answer:

10 whole boards

Step-by-step explanation:

1. 36 divided by 3.75 = 9.6

2. Round to the nearest whole number

3. 9.6 rounded = 10

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Solve the system by substitution.<br> 4y =<br> 5x -10y =-50

Trava [24]

Answer:

$x = -20 ; y =-5$

Step-by-step explanation:

Eqn. 1 ----> 4y = x

Eqn. 2 ----> 5x-10y = -50

(Simplifying eqn.2 further)

$=>5(x-2y)=-50$

$=>x-2y=\frac{-50}{5} =-10$

(Substituting the value of x from eqn. 1)

$=>4y-2y=-10$

$=>2y=-10$

$=>y =\frac{-10}{2} =-5$

Now, substituting the value of y in eqn. 1 ,

$x = 4*(-5)=-20$

7 0

2 years ago

The area of a circular place is 45 square inches. Estimate the radius to the nearest integer.

aksik [14]

A=pi(r)^2
45=pi(r)^2 divide by pi (3.14)
14.33=r^2 square root both sides
3.79=r round
r=4

5 0

3 years ago

Deanna gets 3 problems incorrect on a math quiz. Her score is 85%. How many questions are on the quiz?

Vilka [71]

Answer:

20 questions.

Step-by-step explanation:

Let x be the number of questions on the quiz.

85 % means 85 out of 100 (or 85/ 100 )

85/ 100 = 3/x

--->100-85=15

--->15/3=5

--->100/5=20

Meaning each question was worth 5% and there were 20 questions.

5 0

3 years ago

In the year 2019, Lee County used approximately 55 million gallons of water. Historical trends show that each year the water dem

Harlamova29_29 [7]

Answer:

Approximately 62 gallons of water will be used by Lee County by the year 2025.

Step-by-step explanation:

The amount, in millions gallons of water, spent by Lee County, in t years after 2019, is given by the following equation:

$P(t) = P(0)(1 + r)^{t}$

In which P(0) is the amount spent on 2019 and r is the increse rate, as a decimal.

In the year 2019, Lee County used approximately 55 million gallons of water.

This means that $P(0) = 55$

Historical trends show that each year the water demand is increasing by a rate of approximately 2.1%.

This means that $r = 0.021$

So

$P(t) = P(0)(1 + r)^{t}$

$P(t) = 55(1 + 0.021)^{t}$

$P(t) = 55(1.021)^{t}$

Approximately how many million gallons of water will be used by Lee County by the year 2025?

2025 is 2025 - 2019 = 6 years after 2019. So this is P(6).

$P(t) = 55(1.021)^{t}$

$P(6) = 55(1.021)^{6}$

$P(6) = 62.3$

Rounded to the nearest millions

Approximately 62 gallons of water will be used by Lee County by the year 2025.

3 0

3 years ago

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

41% of U.S. adults have very little confidence in newspapers.

This means that $p = 0.41$

You randomly select 10 U.S. adults.

This means that $n = 10$

(a) exactly five

This is $P(X = 5)$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087$

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480$

$P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125$

$P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019$

$P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001$

Then

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834$

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

(c) less than four.

This is:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051$

$P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355$

$P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111$

$P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058$

So

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575$

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0

3 years ago