We can add or subtract integer. We can also subtract like terms.
<h3>
Answer:</h3>
After 16 downloads the charges will equal each other.
<h3>
Step-by-step explanation:</h3>
First set up an equation
x = Downloads
12 + 1x = 5 + 1.5x
Solve
12 + 1x = 5 + 1.5x <- Rewrite your equation
-5 -5 <- Subtract 5 from both sides
8 + 1x = 1.5x <- Simplify
-1x -1x <- Subtract 1 from both sides
8 = 0.5x <- Simplify
8 = 1/2x <- Convert 0.5 to a fraction to make the next step easier
16 = x <- Multiply both sides by 2 to divide by
a fraction
Answer: Check if numbers are prime. Composite numbers prime factorization (decomposing, breaking numbers down to prime factors). Inscribe them as a product of prime factors, in exponential notation.
Answer:
i) Probability that both candidates employed are women = 5/14
ii) Probability that the second candidate is a woman = 5/8
iii) Probability that the first candidate is a woman given that second one is a woman = 4/5
Step-by-step explanation:
Let the probability that a man is employed be P(M) = 3/8
Probability that a woman is employed P(W) = 5/8
a) Probability that both candidates employed are women = (5/8) × (4/7) = 5/14
b) Probability that the second candidate is a woman = (probability that first candidate is a man and second candidate is a woman) + (probability that first candidate is a woman & second candidate is a woman)
= (3/8)(5/7) + (5/8)(4/7) = (15/56) + (20/56) = 35/56 = 5/8
c) Probability that the first candidate is a woman given that second one is a woman
Given that the second candidate was a women, means that the first candidate-women was selected among other four women.
Probability = (4/8)/(5/8) = 4/5
Answer:
The claim that the scores of UT students are less than the US average is wrong
Step-by-step explanation:
Given : Sample size = 64
Standard deviation = 112
Mean = 505
Average score = 477
To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.
Solution:
Sample size = 64
n > 30
So we will use z test
Formula : 
Refer the z table for p value
p value = 0.9772
α=0.05
p value > α
So, we accept the null hypothesis
Hence The claim that the scores of UT students are less than the US average is wrong
