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LuckyWell [14K]

3 years ago

11

A rental car company charges $62.50 per day to rent a car and $0.07 for every mile driven. Khadija wants to rent a car, knowing

that:
She plans to drive 125 miles.
She has at most $340 to spend.

Write and solve an inequality which can be used to determine
x
x, the number of days Khadija can afford to rent while staying within her budget.

1 answer:

Kamila [148]3 years ago

6 0

62.50 x (0.07 x 125)
0.07 x 125=8.75
62.50 + 8.75=71.25
She can afford it.

I am sure, about the answer but not the equation

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Could somebody help me understand this?

pantera1 [17]

Answer:

M=a2-a1/b2-b1

Step-by-step explanation:

-3-a/-2-1

-4a/3

/-4/-4

a=3/-4

7 0

2 years ago

A number when divided by 780 gives remainder 38 .What

baherus [9]

Given:

A number when divided by 780 gives remainder 38.

To find:

The reminder that would be obtained by dividing same number by 26.

Solution:

According to Euclis' division algorithm,

$a=bq+r$ ...(i)

Where, q is quotient and $0\leq r$ is the remainder.

It is given that a number when divided by 780 gives remainder 38.

Substituting $b=780,\ r=38$ in (i), we get

$a=(780)q+38$

So, given number is in the form of $780q+38$ , where q is an integer.

On dividing $780q+38$ by 26, we get

$\dfrac{780q+38}{26}=\dfrac{780q}{26}+\dfrac{38}{26}$

$\dfrac{780q+38}{26}=30q+\dfrac{26+12}{26}$

$\dfrac{780q+38}{26}=30q+\dfrac{26}{26}+\dfrac{12}{26}$

$\dfrac{780q+38}{26}=30q+1+\dfrac{12}{26}$

Since q is an integer, therefore (30q+1) is also an integer but $\dfrac{12}{26}$ is not an integer. Here 26 is divisor and 12 is remainder.

Therefore, the required remainder is 12.

4 0

3 years ago

What is the difference of the fractions? Use the number line to help find the answer.

vova2212 [387]

<h2>Explanation:</h2>

Hello, remember you need to write complete questions in order to get good and exact answers. Here you haven't provided any fractions, so I'll give you my own fractions.

The first fraction is:

$\frac{1}{2}$

The second fraction is:

$\frac{1}{4}$

So let's say that difference is:

$\frac{1}{2}-\frac{1}{4}$

Therefore, the result is:

$\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

The representation of this problem is shown using the number line below. As you can see, we have written both 1/2 and 1/4 and the difference is also indicated giving the result 1/4. That is, if we walk from 1/4 to 1/2 we'll walk 1/4 units.

6 0

3 years ago

Read 2 more answers

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

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.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

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

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

Assume that ∠A is an acute angle. If sin A = 0.994, use a calculator to find the measure of ∠A to two decimal places. The mea

Sedbober [7]

Basically, you are looking for the ARC SINE of .994 which is
83.72 Degrees
You could also use an online calculator:
http://www.1728.org/trigcalc.htm

8 0

3 years ago