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3 years ago

12

Nico, Aditi, Erick, Raj, and Sandra are solving the equation. Two-fifths (5 + x) = 8 Each student begins to solve the problem as

shown.

2 answers:

Crank3 years ago

8 0

Answer:A,D,E

Step-by-step explanation:

nico raj, n sandra

Anit [1.1K]3 years ago

4 0

Answer:Nico Raj & Sandra hope this help x3

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PLEASE HELP ME!! 99 POINTS!!!!!!

barxatty [35]

Answer:

=> 51b - 24

=> 51(4) - 24

=> 204 - 24

=> 180

8 0

3 years ago

Read 2 more answers

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

3 years ago

Let f(x)=(x+4)^2−3.

Alexxx [7]

If you start with f(x)
f(x) = (x+4)^2-3
and add 9 to both sides, then we get
f(x)+9 = (x+4)^2-3+9
f(x)+9 = (x+4)^2+6
f(x)+9 = g(x)
g(x) = f(x)+9
this shows that g(x) is the translation of f(x) up 9 units

Answer choice C

8 0

4 years ago

Select the correct answer from each drop-down menu.

Oksi-84 [34.3K]

Answer:

$we \: have \: ohms \: law \: h = {i}^{2} rt \\ \frac{h}{t} = {i}^{2} rt \div t = p \\ p = {i}^{2} r....(1) \\ we \: have \: v = ir \\ i = \frac{v}{r} .....(2) \\ put \: (2) \: in \: (1) \\ then \: p = { (\frac{v}{r}) }^{2} r \\ p = \frac{ {v}^{2} }{r} \\ v = \sqrt{pr} = {(pr)}^{ \frac{1}{2} } \\ 2) \\ here \: r = 32 \: and \: p = .5 \\ then \: v = \sqrt{32 \times .5} \\ = \sqrt{16} = 4volt \\ thank \: you$

6 0

3 years ago

I THINK IT"S "D"

Ainat [17]

Slope m=(y2-y1)/(x2-x1)=(10-4)/(4-2)=6/2=3

y = mx+b
y=3x+b
Take point (2,4),
4=3*2+b
4=6+b
b=-2
y=3x-2
Answer is D. <span>y = 3x − 2</span>

6 0

3 years ago