Plz help!! tell me which ones you solve

Mathematics

1 answer:

Sauron [17]3 years ago

7 0

Answer:

Step-by-step explanation:

17)-8(5-n)=-96

-8*(5-n)-(-96)=0

(0 - 8 • (5 - n)) - -96 = 0

8n + 56 = 8 • (n + 7)

8 • (n + 7) = 0

Solve : 8 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solve : n+7 = 0

Subtract 7 from both sides of the equation :

n = -7

n = -7 is your answer

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Select all instances in which the variable described is binomial.

tino4ka555 [31]

Answer: a. , c. , d., e.

Step-by-step explanation:

A variable that counts how many times a certain event occurs in a particular number of trials is known as binomial random variable.

For each trial, there exist only two outcomes .

The probability of for each event is the same on each trial.

a. Event has two outcomes with same probability as 0.50, therefore the random variable represents the total number of flips required to get tails is a binomial random variable.

b. Total guidelines are 5.

Here total outcomes are not 2 , it does not meet with the conditions of binomial.

c. The random variable represents the total number of children from this pair of parents with blue eyes has two outcomes (where has or not.)

also, the probability of having blue eyes is same in each trial, so it represents binomial random variable.

d. The random variable represents the total number out of 567 customers with a checking account has two outcomes (checking or savings).

So, it represents binomial random variable.

e. The random variable represents the total number of ace cards observed has two outcomes ( ace or not ace).

So it represents the binomial random variable.

7 0

2 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$