For the function f(x)=x^2-5x+6, where is the y-intercept?

there are 9000 children in a town the probability that a child is affected by a virus 1/3 how many children are not likely to be

Sophie [7]

Answer:

6,000 children

Step-by-step explanation:

1/3 of 9,000 is 3,000

Meaning that out of 9,000 children, 3,000 WILL get affected. Since we are trying to figure out how many children will NOT get affected, just subtract the number of infected by the total amount which would be 6,000

7 0

2 years ago

Jenny was planning a trip to the United Arab Emirates. Before going, she did some research and learned that the exchange rate is

Blizzard [7]

4 dirhams is equal to 1 dollar so if you were to multiply each by 5 you would have 20 dirhams equaling 5 dollars

5 0

2 years ago

Evaluate the following expressions if a = 2, b = 3, x = 4, and y = 5.<br> b^2+3(2x-y)

butalik [34]

Hi there! Hopefully this helps!

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<h2 /><h2>Answer: 18</h2><h2>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</h2><h2 />

First we need to substitute the value of the variable into the expression.

$b^{2}+3(2x-y)$ = $3^{2} + 3 (2(4)-5)$ .

Now we need to solve $3^{2} + 3 (2(4)-5)$ .

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$3^{2} + 3 (2(4)-5)$

First, calculate 3 to the power of 2 and get 9.

$9+3(2\times 4-5)=18$

Multiply 2 and 4 to get 8.

$9+3(8-5) =18$

Subtract 5 from 8 to get 3.

$9+3\times 3 =18$

Multiply 3 and 3 to get 9.

$9+9 =18$ .

<h2>And the answer is, you guessed it, 18!</h2>

7 0

2 years ago

Find the distance between the two points, round to the nearest tenth if needed. (-6,-2) and (1, 6)

Gnoma [55]

Due to the distance formula the answer should be root 113 or 10.63

7 0

3 years ago

A veterinarian wants to know if pit bulls or golden retrievers have a higher incidence of tooth decay at the age of three. The v

BigorU [14]

Answer: B) [-0.0332,0.1332]

Step-by-step explanation: <u>Confidence</u> <u>Interval</u> is an interval where we can be a percentage sure the true mean is.

The confidence interval for a difference in population proportion is calculated following these steps:

First, let's find population proportion for each population:

$p_{1}=\frac{30}{120}=0.25$

$p_{2}=\frac{32}{160}=0.2$

Second, calculate standard deviation for each proportion:

$\sigma_{1}=\sqrt{\frac{0.25(0.75)}{120} } = 0.0395$

$\sigma_{2}=\sqrt{\frac{0.2(0.8)}{160} } = 0.0316$

Now, we calculate standard error for difference:

$SE=\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}$

$SE=\sqrt{0.0395^{2}+0.0316^{2}}$

SE = 0.0505

The z-score for a 90% CI is 1.645.

Then, confidence interval is

$p_{1}-p_{2}$ ± z-score.SE

$0.25-0.2$ ± $1.645(0.0505)$

0.05 ± 0.0831

The limits of this interval are:

inferior: 0.05 - 0.0831 = $-0.0332$

superior: 0.05 + 0.0831 = 0.1332

The 90% confidence interval for the difference in the population proportion of pit pulls and golden retrievers is $[-0.0332,0.1332]$ .

8 0

3 years ago