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3 years ago

If x represents a number then write an expression for a number that is three more than one third the value of x

Mathematics

2 answers:

oksano4ka [1.4K]3 years ago

8 0

Answer:

1/3x+3

It is asking for 1/3 of x which is 1/3x

three more essentially means add 3

So 1/3x+3 is the answer

Katyanochek1 [597]3 years ago

6 0

Answer:

1/3x+3, because 1/3 times x is one third the value of x, and the "+3" describes the 3 more.

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A stunt man wants to drive a car off a ramp at an angle of 35∘ from the ground. He already has a ramp with an angle of 15∘. He p

eduard

Answer:

The second ramp needs to have 25°.

Step-by-step explanation:

Required ramp angle for driving the car from the ground = 35°.

First ramp angle = 15°

Therefore, second ramp angle equals required ramp angle minus first ramp angle, which is = 35° - 15° = 20°

The 20° ramp angle will make the total ramp angle to be equal to 35°.

That is 15° + 20° = 35°

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3 years ago

I need the equation of the line in the graph

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4 years ago

For my math it says someone ordered a poster with a height of 2 feet and width of 0.7 feet what is the area of the poster

Phantasy [73]

STEP-BY-STEP SOLUTION:

Let's solve this problem step-by-step.

As posters are generally rectangular in shape, we will establish the formula for the area of a rectangle which we will be using as displayed below:

Area = height × width

Now let's find the area of the postwr using the above formula.

Height = 2 feet

Width = 0.7 feet

Area = height × width

Area = 2 × 0.7

Area = 1.4 square feet

ANSWER:

Therefore, the answer is:

The area of the poster is 1.4 square feet.

Please mark as brainliest if you found this helpful! <3
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3 years ago

5 cups to 8 cups identify the percent

Kipish [7]

Well I am going to guess and i don't know if it right but 156.25ml

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3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago