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2 years ago

PLEASE HELP ASAP I WILL MARK BRAINLIST (Questions and Answers pictured)

Mathematics

2 answers:

Strike441 [17]2 years ago

8 0

Answer:

g(x)=3f(2x)

I think

nexus9112 [7]2 years ago

6 0

Answer:

The first equation, sorry can’t explain.

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8 × 10 + 6 − 2 7 + 5 × 8 − 4

riadik2000 [5.3K]

Answer:

Step-by-step explanation:

(8)(10)+6−27+(5)(8)−4

=80+6−27+(5)(8)−4

=86−27+(5)(8)−4

=59+(5)(8)−4

=59+40−4

=99−4

=95

hope this helped :)

6 0

2 years ago

Read 2 more answers

ASAP I NEED HELP PLZ PLZ PLZ PLZ

liberstina [14]

Answer: 64 ................ :)

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A basketball player has made 70% of his foul shots during the season. If he shoots 3 foul shots in tonight's game, what is the

yulyashka [42]

Answer:

There is a 34.3% probability that he makes all of the shots.

Step-by-step explanation:

For each foul shot that he takes during the game, there are only two possible outcomes. Either he makes it, or he misses. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 3, p = 0.7$

What is the probability that he makes all of the shots?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.7)^{3}.(0.3)^{0} = 0.343$

There is a 34.3% probability that he makes all of the shots.

7 0

3 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

Find the area under the standard normal curve between z = −2.57 and z = 1.27. Round your answer to four decimal places, if neces

dexar [7]

P(-2.57 ≤ Z ≤ 1.27) = P(Z ≤ 1.27) - P(Z ≤ -2.57)

P(Z ≤ 1.27) = 0.8980
P(Z ≤ -2.57) = 1 - P(Z ≤ 2.57) = 1 - 0.99488 = 0.00512

P(-2.57 ≤ Z ≤ 1.27) = 0.8980 - 0.00512 = 0.89288
0.8929 (4 d.p.)

4 0

3 years ago

*PLEASE HELP ASAP I WILL MARK BRAINLIST* (Questions and Answers pictured)

PLEASE HELP ASAP I WILL MARK BRAINLIST (Questions and Answers pictured)