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3 years ago

Look at the picture. Which one is he answer?

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

6 0

A (2^2/2^-1)is the correct answer because 2^-5 * 2^8 = 8 and 2^2/2^-1 = 8.

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My Someone please explain why G is 125 and H is 62!

MAVERICK [17]

I don’t know why but i’m first so pleaseeee

Step By Step Explanation:

7 0

3 years ago

The least common denominator of two fractions is 28. If you subtract the two denominators, their difference is 10. What are the

wel

14 and 28 are two different common denominators for the fractions and we convert both fractions to eighteenths we can subtract 3 eighteenths from 10

so what do you think the answer is
this is just a example

7 0

3 years ago

What should I buy? A study conducted by a research group in a recent year reported that of cell phone owners used their phones i

Llana [10]

Answer:

The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

Step-by-step explanation:

The question is incomplete:

What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions?

We can model this as a binomial random variable, with p=0.57 and n=14.

$P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}$

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

$P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\$

$P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\$

$P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\$

$P(x\geq7)=0.1824+0.2115+0.1869+0.1239+0.0597+0.0198+0.004+0.0004\\\\P(x\geq7)=0.7886$

7 0

3 years ago

At Andrew Jackson High School, students are only allowed to enroll in AP U.S. History if they have already taken AP World Histor

o-na [289]

Answer:

About 0.22

Explanation:

There are two sets:

Set W of incoming seniors who took AP World History, and

Set E of incoming seniors who took AP European History

And there is a subset, which is the intersection of those two sets:

Subset W ∩ E of senior students who took both.

The incoming seniors who are allowed to enroll in AP U.S. History, call them the subset S, is the set of those students that belong to W or E or both W ∩E.

By property of sets:

S = W + E - W∩E = 175 + 36 - 33 = 178

Then, 178 out of 825 incoming seniors took one or both courses, and the desired probability of a randomly selected incoming senior is allowed to enroll in AP U.S. History is:

178/825 = 0.21576 ≈ 0.22

6 0

3 years ago

Explain and solve . hhhhgghgfghhg

Aleks [24]

Answer:

mark as brainliest and drop some thanks!!!

5 0

3 years ago