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scoundrel [369]
3 years ago
6

Jacob needs less than 5 C's on his transcript to qualify for UGA. He already has 1 C. At most, how many more C's can he get and

still be able to go to UGA? Write an inequality and solve. A) x + 1 ≤ 5; x ≤ 4 B) x + 1 ≥ 5; x ≥ 4 C) x + 1 < 5; x < 4 D) x + 1 > 5; x > 4
Mathematics
2 answers:
svlad2 [7]3 years ago
7 0

Answer:

C. x + 1 < 5; x < 4

Step-by-step explanation:

Since he already got 1 C, he can only have 3 more. Since x has to be <em>less than 4</em>, that rules out B and D instantly. A is ruled out because x can't be equal to 4 since you need less than 5 C's to get into UGA.


Hope this helps. :)

gtnhenbr [62]3 years ago
6 0

Answer:

The answer is c

Step-by-step explanation:

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4 years ago
A 28 ft ladder is leaned against the side of a building and its base is 5 ft away from the building. How far up the building wil
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Answer:

27.55ft

Step-by-step explanation:

28²-5²=x

784-25=759

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3 years ago
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4 0
3 years ago
Read 2 more answers
Construct the discrete probability distribution for the random variable described. Express the probabilities as simplified fract
lisov135 [29]

Answer:

P(X = 0) = 0.03125

P(X = 1) = 0.15625

P(X = 2) = 0.3125

P(X = 3) = 0.3125

P(X = 4) = 0.15625

P(X = 5) = 0.03125

Step-by-step explanation:

For each toss, there are only two possible outcomes. Either it is tails, or it is not. The probability of a toss resulting in tails is independent of any other toss, which means that the binomial probability distribution is used to solve this question.

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 combinations 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.

Fair coin:

Equally as likely to be heads or tails, so p = 0.5

5 tosses:

This means that n = 5

Probability distribution:

Probability of each outcome, so:

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

P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125

P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625

P(X = 2) = C_{5,2}.(0.5)^{2}.(0.5)^{3} = 0.3125

P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125

P(X = 4) = C_{5,4}.(0.5)^{4}.(0.5)^{1} = 0.15625

P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125

5 0
3 years ago
find the measure of the arc or angle indicated. assume that lines which appear to be diameters are actual diameters just enter t
konstantin123 [22]

Answer:

I think D

if wrong correct me plssssssss

have a nice day

And good luck if you have exam

#Captainpower

8 0
3 years ago
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