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Alenkasestr [34]

3 years ago

10

Simplify (2xy^3)^4 A.8xy12 B.2x4y12 C.16xy12 D.16x4y12

2 answers:

suter [353]3 years ago

6 0

Answer:

D) $(16)*x^{4} *(y^{12})$ is correct answer .

Step-by-step explanation:

Given : $(2xy^{3})^{4}$ .

To find : Simplify.

Solution : We have given that $(2xy^{3})^{4}$ .

By the exponential rule : $(ab)^{c} = a^{c}*b^{c}$ .

$(2xy^{3})^{4}$ = $(2^{4})*x^{4} *(y^{3})^{4}$ .

$(2xy^{3})^{4}$ = $(16)*x^{4} *(y^{12})$ .

Therefore, D) $(16)*x^{4} *(y^{12})$ is correct answer .

Anna35 [415]3 years ago

3 0

The answer is D if you would simplify this (2xy^3)^4

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Step-by-step explanation:

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6(-3-7)=? Please help!

katovenus [111]

Answer:

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Step-by-step explanation:

Distribute

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If my answer is right, please give it the brainliest.

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3 years ago

Read 2 more answers

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

So

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

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3 years ago

I am working on a test review. How can I write -1 <= 2x+7 <21 as two inequality statements? How do I join them with and/or

Tanya [424]

Answer:

$( - 1 > 2x + 7)(2x + 7 < 21)$

Not sure about the and/or statement

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3 years ago