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3 years ago

I need help on question 35

Mathematics

1 answer:

Novay_Z [31]3 years ago

5 0

Pretty sure it's everything except irrational

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Find f (2b^2) for (x)=x^2-4x

BlackZzzverrR [31]

Answer:

Step-by-step explanation:

We are given the function:

$f(x) = x^2 - 4x$

And we want to find:

$\displaystyle f(2b^2)$

Substitute:

$\displaystyle f(2b^2) =(2b^2)^2 -4(2b^2)$

And evaluate:

$\displaystyle \begin{aligned} f(2b^2) &=(2b^2)^2 -4(2b^2) \\ &=(4b^4) +(-8b^2) \\ &= 4b^4 - 8b^2 \end{aligned}$

In conclusion, our answer is A.

7 0

2 years ago

A pharmacist has 40% and 60iodine solutions on hand. How many liters of each iodine solutions will be required to produce 4 lite

Schach [20]

A pharmacist has 40% and 80% of iodine solutions on hand. How many liters of each iodine solution will be required to produce 4 liters of a 50% iodine mixture?

Let x = liters of 40% iodine

then

4-x = liters of 80% iodine

Using algebra:

.40x + .80(4-x) = .50(4)

.40x + 3.20-.80x = 2

3.20-.40x = 2

x = 4 liters (40% iodine)

80% iodine:

4-x = 4-4 = 0 liters needed (80% iodine)

4 0

3 years ago

HELP ASAP

lora16 [44]

I solved it on a piece of paper. I can not answer why I got the same answer, but I hope my method helped

3 0

2 years ago

Which situation could be solved using the equation-4+4=0

Taya2010 [7]

A change in temperature.

3 0

2 years ago

A recent study found that the average length of caterpillars was 2.8 centimeters with a

pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 2.8, \sigma = 0.7$ .

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4 - 2.8}{0.7}$

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

4 0

2 years ago