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3 years ago

"The sum of 4 and a number, divided by 2 is less than -6."

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

3 0

Answer:

$\frac{4+n}{2} < -6$ , n < -16

Step-by-step explanation:

To write this into an inequality you need to recognize keywords. The first keyword is "sum" this means the two numbers must be added; so the first part of the inequality is 4+n. Then, it says the sum is "divided by" 2; this means both numbers must be put over two like $\frac{4+n}{2}$ . Finally, this is "less than" negative. Therefore, this must be an inequality with a sign opening towards -6. So the final answer is $\frac{4+n}{2} < -6$ .

To solve the inequality isolate the variable. Multiply both sides by 2, so the new inequality is $4+n$ . Next, subtract 4 from both sides. The answer is n < -16.

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4.20. According to a report released by the National Center for Health Statistics, 51% of U.S. households has only cell phones (

Sergeu [11.5K]

Answer:

The probability that the household has only cell phones and has high-speed Internet is 0.408

Step-by-step explanation:

Let A be the event that represents U.S. households has only cell phones

Let B be the event that represents U.S. households have high-speed Internet.

We are given that 51% of U.S. households has only cell phones

P(A)=0.51

We are given that 70% of the U.S. households have high-speed Internet.

P(B)=0.7

We are given that U.S. households having only cell phones, 80% have high-speed Internet. A U.S household is randomly selected.

P(B|A)=0.8

$\frac{P(A\capB)}{P(A)}=0.8\\P(A\capB)=0.8 \times P(A)\\P(A\capB)=0.8 \times 0.51\\P(A\capB)=0.408$

Hence the probability that the household has only cell phones and has high-speed Internet is 0.408

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2 years ago

Let X1 and X2 be two random variables following Binomial distribution Bin(n1,p) and Bin(n2,p), respectively. Assume that X1 and

ryzh [129]

Answer:

a) X1+X2 have distribution Bi(n1+n2, p)

P(X1+X2 = 1 | X2 = 0) = np(1-p)ⁿ¹⁻¹

P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹

P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

Step-by-step explanation:

Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).

Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).

If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹

If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹

Thus,

P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

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Step-by-step explanation:

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"The sum of 4 and a number, divided by 2 is less than -6."​

"The sum of 4 and a number, divided by 2 is less than -6."