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2 years ago

Can someone please help me on this? thanks!!

Mathematics

1 answer:

tankabanditka [31]2 years ago

6 0

Answer:

store C is the best with .122 per ounce the others are .129 and .135

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ΔEʹFʹGʹ is a dilation of ΔEFG with the origin as the center of dilation.

vekshin1

the answer a is 6 and b is 2

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3 years ago

Simplify 8x + 7 – 2x – 4 completely.

harkovskaia [24]

Answer:

6x+3

Step-by-step explanation:

8x +7 - 2x - 4

8x-2x+7-4 ( combine like terms)

answer:

6x+3

Reminder:Whenever combining like terms, the sign changes the same

Hope this helps and make sure to thank me!

6 0

2 years ago

Read 2 more answers

For the graph, which are possible functions of f and g?

GuDViN [60]

Answer:

f= 3ˣ

g= (1/3)ˣ

Step-by-step explanation:

5 0

2 years ago

If you if you take away 25% of the radius of the suspension how much raise would you have left over of the hundred percent you t

natulia [17]

Answer:

Step-by-step explanation:

25% of 100 is 25

100 - 25 = 75

Give me brainllest

3 0

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 <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> 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

2 years ago