I need help on this

`ANSWER ASAP GIVING BRAILIEST AND STUFF!

kykrilka [37]

Answer:

If the dimension increased by doubled, tripled or quadrupled, then it will be 2x, 3x, or 4x for each dimension.

3 0

3 years ago

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Which equation is the equation of the line, in point-slope form, that has a slope of 2 and passes through the point (-8, 1)?

lutik1710 [3]

Answer:

y=2x+17

Step-by-step explanation:

y-y1=m(x-x1)

y-1=2(x-(-8))

y-1=2(x+8)

y=2x+16+1

y=2x+17

5 0

2 years ago

The conservation club has 32 members. There are 18 girls in the club.

leva [86]

Answer: 9/7

Step-by-step explanation:

Girls = 18

Total = 32

32-18 = 14

18/14 simplify

= 9/7 (B)

8 0

2 years ago

11(2x + 1) = 4x - 7 show work

Harlamova29_29 [7]

Answer:

0.22

Step-by-step explanation:

22x + 11 = 4x - 7

22x - 4x = -7 + 11

18x = 4

x = 4/18

x = 0.22

Please mark as brainliest.

8 0

3 years ago

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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