Please solve for the maximum value f(x) = -2x^2 - 12x

A book 500 pages (500 sheets of paper) long is 2.25 inches thick. If each cover is 0.125 inches thick, how thick is each sheet o

luda_lava [24]

Subtract the thickness of the covers from the total thickness of the 500-pages book. Since each cover is 0.125 inches thick, double this value and subtract from 2.25 inches. This results to 2 inches. To determine the thickness of each of the pages, divide 2 inches by 500 pages. Therefore, the thickness of a single page is 4x10^-3 inch.

6 0

3 years ago

Write an equation that could be used to find the measure of angle A

Ainat [17]

Answer: C

Step-by-step explanation: The numerator is the angle measure, and the denominator is the side length. For angle B, the angle is 47 degrees. The opposite side is b, which is 85. We are finding the angle A, which is the numerator. The side 94 is opposite of angle A.

7 0

3 years ago

a cooler contains 50 bottles; 30 bottles of water, which 8 are lemon flavored and 20 bottles of tea, of which 5 are lemon flavor

xenn [34]

Answer:

The required probability is $\dfrac{13}{50}$

Step-by-step explanation:

Total number of bottles = 50

Total number of water bottles = 30

Total number of lemon flavored water bottles = 8

Total number of tea bottles = 20

Total number of lemon flavored tea bottles = 5

Probability of selecting a lemon flavored water bottle = Probability of selecting a water bottle $\times$ Probability of selecting a lemon bottle out of water bottles.

Probability of selecting a lemon flavored tea bottle = Probability of selecting a tea bottle $\times$ Probability of selecting a lemon bottle out of tea bottles.

Formula for probability of an event E can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

Probability of selecting a water bottle:

$\dfrac{30}{50} = \dfrac{3}{5}$

Probability of selecting a lemon flavored bottle from water bottle:

$\dfrac{8}{30}$

Probability of selecting a lemon flavored water bottle = P(A)

$\dfrac{3}{5} \times \dfrac{8}{30} = \dfrac{4}{25}$

Probability of selecting a tea bottle:

$\dfrac{20}{50} = \dfrac{2}{5}$

Probability of selecting a lemon flavored bottle from tea bottle:

$\dfrac{5}{20} = \dfrac{1}{4}$

Probability of selecting a lemon flavored tea bottle = P(B)

$\dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{1}{10}$

The required probability is:

P(A) + P(B):

$\dfrac{4}{25} + \dfrac{1}{10}\\\Rightarrow \dfrac{8+5}{50}\\\Rightarrow \dfrac{13}{50}$

8 0

3 years ago

A local garden contains 1/12 lettuce, 3/12 carrots, 4/12 of corn and the rest red radishes. How much of the garden is radishes?

melamori03 [73]

Answer:

The answer to your question is a) 4/12 b) 5/12

Step-by-step explanation:

Data

Lettuce = 1/12

Carrots = 3/12

Corn = 4/12

Red radishes = ?

a)

The total of the garden is 12/12

Red radishes = 12/12 - 1/12 - 3/12 - 4/12

= 12/12 - 8/12

= 4/12

4/12 of the garden are red radishes.

b)

Corn of lettuce = 1/12 + 4/12

= 5/12

5/12 of the garden are corn or lettuce.

3 0

3 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$