P(6-8 years old) When answering the question I put 1/5 could you tell me what I did wrong?

According to the graph what is the value of the constant in the equation below

attashe74 [19]

Answer:

I don't know the answer. .

3 0

2 years ago

If x – 9 is a factor of x2 – 5x – 36, what is the other factor a. x – 4 b. x 4 c. x – 6 d. x 6

schepotkina [342]

X^2 -5x -36 = x*x -9x+4x -9*4 = x(x-9) +4(x-9) = (x+4)(x-9)

so the second other factor is x+4 so hope that choice wan being x+4 than this will be right just you missed the sign pf plus

hope this will help you

5 0

3 years ago

Read 2 more answers

using the 28/36 ratio, determine the maximum allowable recurring debt for someone with an annual income of $86,250.

murzikaleks [220]

Given:
28/36 ratio
annual income: 86,250

28+36 = 64

28/64 * 86,250 = 0.4375 * 86,250 = 37,734.375
36/64 * 86,250 = 0.5625 * 86,250 = 48,515.625

28/36 ⇒ 37,734.375/48,515.625

5 0

3 years ago

Is 85/95 greater than 64/76

elena-14-01-66 [18.8K]

The answer is yes I believe

5 0

3 years ago

For a certain candy, 20% of the pieces are yellow, 15% are red, 20% are blue, 20% are green, and the rest are brown. a

OLEGan [10]

Answer:

Step-by-step explanation:

Based on the question we are given the percentages of each of the types of candies in the bag except for brown. Since the sum of all the percentages equals 75% and brown is the remaining percent then we can calculate that brown is (100-75 = 25%) 25% of the bag. Now we can show the probabilities of getting a certain type of candy by placing the percentages over the total percentage (100%).

Brown: $\frac{25}{100}$
Yellow or Blue: $\frac{20}{100} +\frac{20}{100} = \frac{40}{100}$ ....add the numerators

Not Green: $\frac{80}{100}$ .... since the sum of all the rest is 80%

Stiped: $\frac{25}{100}$ .... there are 0 striped candies.

Assuming the ratios/percentages of the candies stay the same having an infinite amount of candy will not affect the probabilities. That being said in order to calculate consecutive probability of getting 3 of a certain type in a row we have to multiply the probabilities together. This is calculated by multiplying the numerators with numerators and denominators with denominators.

3 Browns: $\frac{25*25*25}{100*100*100} = \frac{15,625}{1,000,000} = \frac{1.5625}{100}$

the 1st and 3rd are red while the middle is any. We multiply 15% * (total of all minus red which is 85%) * 15% like so.

$\frac{15*85*15}{100*100*100} = \frac{19,125}{1,000,000} = \frac{1.9125}{100}$

None are Yellow: multiply the percent of all minus yellow three times.

$\frac{80*80*80}{100*100*100} = \frac{512,000}{1,000,000} = \frac{51.2}{100}$

At least 1 green: multiply the percent of green by 100% twice, since the other two can by any

$\frac{20*100*100}{100*100*100} = \frac{200,000}{1,000,000} = \frac{20}{100}$

4 0

3 years ago