All categories

3 years ago

at an amusement park, a person spends $30 on admission and food, and then goes on r number of rides that cost $2 each

1 answer:

3 0

30 dollars would be your initial amount which would be the b part of your equation, y=mx +b. 2r would be the mx part. Your equation would read y=2r+30 and then you would plug in how many rides they went on into the r space to find y

You might be interested in

What is the effect on the graph of the function f(x) = 2x when f(x) is replaced with f(x − 3)? A) translate vertically 3 units u

Elza [17]

Answer:

the answer is D, translate horizontally 3 unit right.

5 0

3 years ago

A vial contains 20mL of medicine.If each dose is 1/3 of the vial,how many mL is each dose.Express your answer as a decimal

Genrish500 [490]

Answer:

each dose = 6.7 mL ( to 1 decimal place)

Step-by-step explanation:

Total volume of vial = 20mL

each dose of vial = 1/3 of the total volume

let us express this fraction as a real number:

1/3 of 20 = 1/3 × 20

Since we are to express the answer as a decimal, let us convert the fraction to decimal:

1/3 = 0.3333

∴ 1/3 × 20 = 0.3333 × 20 = 6.667 mL

since we are dealing with drugs, it'll be appropraite to express the answer to 1 d.p

∴ each dose = 6.7 mL

4 0

2 years ago

Which of the following are solutions to the equation below? Check all that apply. (3x - 5)2 = 19 O A. x= 19+ olm B. x= 119 +5 I

asambeis [7]

Solution

$(3x-5)^2=19$

For this case we can take square root in both sides and we have:

$3x-5=\pm\sqrt[]{19}$

And solving for x we got:

$x=\frac{5\pm\sqrt[]{19}}{3}$

then the solutions for this case are:

B and E

8 0

7 months ago

The sum of 9 and a number is twice a number plus 14. How would I write this as an equation?

valina [46]

1+2+3+4+5+6+7+8+9×2+14

6 0

2 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

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

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

2 years ago