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3 years ago

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1) Suzanne will pay $14 to get her hair cut. She wants to leave the hairdresser a 15% tip. What is the total cost of the haircut

, including tip?

1 answer:

Kryger [21]3 years ago

7 0

Answer:

$16.10

Step-by-step explanation

14.00x .15=2.10

14.00+2.10=16.10

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All of the following are equivalent except _____. A:5 over 4. B:1.25. C:12.5%. D:125%.

earnstyle [38]

Hello, there!

5/4 = 1.25

So, we know that the answer is not A or B.

1.25 = 125%

Now, we know that the answer is not D ether.

So, the answer must be C.

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Plz Um can someone Please help me I really need help with this

photoshop1234 [79]

You seemed to have asked this multiple times. I have answered one of yours. Its D.

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Jamal bought donuts and cupcakes. He bought three times as many donuts as cupcakes. Donuts cost $0.50 each. Cupcakes cost $1.00

jolli1 [7]

Answer:

Step-by-step explanation:

Let

x= Quantity of donuts

y= Quantity of cupcakes

He bought 3 times as many donuts as cupcakes

Donuts=y= 3x

Donuts=0.50 each

Cupcakes=1.00 each

PxQx + PyQy = 10.00

0.50(x) + 1.00(y) = 10.00

0.50x + 1.00(3x)=10.00

0.50x+3.00x=10.00

3.50x=10.00

Divide both sides by 3.50

x=10.00/3.50

=2.86

y=3x

y=3(2.86)

=8.58

Jamal bought 2.86 donuts and 8.58 cupcakes

Check:

PxQx + PyQy = 10.00

0.50(2.86) + 1.00(8.58) = 10.00

1.43 + 8.58=10.00

10.01=10.00

10.01 approximately 10.00

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3 years ago

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Step-by-step explanation:

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3 years ago

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the name Joe is very common at a school in one out of every ten students go by the name. If there are 15 students in one class,

kumpel [21]

Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.

For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the <em>binomial distribution</em> is used to solve this question.

<h3>Binomial probability distribution </h3>

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

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

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:

One in ten students are named Joe, hence $p = \frac{1}{10} = 0.1$ .

There are 15 students in the class, hence $n = 15$ .

The probability that at least one of them is named Joe is:

$P(X \geq 1) = 1 - P(X = 0)$

In which:

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

$P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059$

Then:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2059 = 0.7941$

0.7941 = 79.41% probability that at least one of them is named Joe.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377

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2 years ago