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3 years ago

If Jamie has five dollars and purchases a soda for one dollar and fifteen cents, how much change does Jamie get back?

Mathematics

1 answer:

yaroslaw [1]3 years ago

3 0

Five dollars is $5.00.
The soda's cost is $1.15.
Simply subtract 5.00 by 1.15 to get
5.00 - 1.15 = $3.85

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The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection.

This means that,

Probability that at least one of them needs replacement at the first inspection =

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

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3 years ago

What can you conclude about the classification of silicon?

Orlov [11]

Answer:

That i will be happy to know that i don't now and silicon is a element

Step-by-step explanation:

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3 years ago

Ali, Ben and Clare each played a game. Clare's score was seven times Ali's score. Ben's score was half of Clare's score. Write d

Lerok [7]

Answer:

The answer is 2 : 7 : 14

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3 years ago

What are the coordinates of the point on the directed line segment from (-4,1) to (1,6) that partitions the segment into a ratio

Aleksandr [31]

Answer:

a segment is partitioned at a ratio of 1:3, then the point is one-fourth of the distance from (-4,-1) to (2,7).

To compute the x-coordinate of that point, you will need to compute one-fourth of the x-distance between 2 and -4 then add it to -4: (2--4)/4 = 1.5; 1.5 + -4 = -2.5.

To compute the y-coordinate of that point, you will need to compute one-fourth of the y-distance between 7 and -1 then add it to -1: (7--1)/4 = 2; 2 + -1 = 1.

The point is (-2.5,1)

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2 years ago

Suppose A and B represent two different school populations where A > B and A and B must be greater than 0. Which of the follo

Ulleksa [173]

We can reject the last one: subtracting a non-zero value will result in a smaller value.

So now we have:

2(A + B)
(A + B)2
A2 + B2

If all of them are mulptiplications, then they are all equivalent!

I mean by this, if what you meant is this:


2*(A + B)
(A + B)*2
A*2 + B*2

If there is no sign, then the multiplication sign is implicit,

and all of these expressions say exactly the same: two of A and two of B.

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3 years ago