If the temperature is at 6 and it rises to 22 what is the temperture

Find the y - intercept of y=3x+15 and then interpret it

oksian1 [2.3K]

Answer:

The y-intercept to the given line equation y=3x+15 is (0,15).

Step-by-step explanation:

Given line equation is y=3x+15

To find the y-intercept from the given line equation y=3x+15:

y-intercept when x=0

That is substitute x=0 in the given line equation

y=3x+15

y=3(0)+15

y=0+15

y=15

Therefore y=5

The point (0,15) on the graph where the line y is intersected is y-intercept.

In other words y-intercept is the point (0,15) where the given line equation y=3x+15 crosses the y-axis.

Therefore the y-intercept to the given line equation y=3x+15 is (0,15).

5 0

3 years ago

Hhhhhhhhhhhheeeeeeeelllllllllllpppppppppppp me

Anettt [7]

I might be wrong but im pretty sure the answer is B.

4 0

2 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:

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

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