Solve for the value of c.
solve for the value of c. - 1

Answers

Answer 1
Answer: c=15, (Check) 7(15)=105, 105-7=98; 6(15)=90, 90-8=82. 98+82=180

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Write the prime factorization of 20 using exponents

Answers

Prime factorization expresses an integer in terms of multiples of prime numbers. The prime factorization of 20 using exponents can be written as 2²×5¹.

What is prime factorization?

Factorization is expressing a mathematical quantity in terms of multiples of smaller units of similar quantities.

Prime factorization is when all those factors are prime numbers.

Thus, prime factorization expresses an integer in terms of multiples of prime numbers.

The prime factorization of 20 using exponents can be written as,

20 = 1 × 2 × 2 × 5

    = 2² × 5¹

Hence, the prime factorization of 20 using exponents can be written as 2²×5¹.

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Here you go

2 to the second power times 5

On a coordinate graph, points F (4,1), G (4,-1), H (-7,1) and J (-7,-1) are the vertices of a rectangle. What is the area of the rectangle? PLEASE EXPLAIN HOW YOU GOT THE ANSWER!!!!

Answers

The sides of the rectangle are aligned with the coordinate axes, so the dimensions of the rectangle are easily determined to be 4 -(-7) = 11 units by 1 -(-1) = 2 units. As with any rectangle, the area is the product of length and width.

The area is 11*2 = 22 square units.

Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of 14 flights is studied. Round the probabilities to at least four decimal places.(a) Find the probability that all 12 of the flights were on time.(b) Find the probability that exactly 10 of the flights were on time.(c) Find the probability that 10 or more of the flights were on time.(d) Would it be unusual for 11 or more of the flights to be on time?

Answers

Analyzing a sample of 14 flights at Denver International Airport, the probability of 10 or more flights arriving on time is 0.3783, and the probability of 11 or more flights arriving on time is 0.2142, which is not considered unusual.

(a) All 12 of the flights were on time.

(b) Exactly 10 of the flights were on time.

(c) 10 or more of the flights were on time.

(d) Would it be unusual for 11 or more of the flights to be on time?

We can use the binomial probability formula to solve this problem. The binomial probability formula is:

P(k successes in n trials) = (n choose k) *(p)^k * (q)^(^n^-^k^)

where:

n is the number of trials

k is the number of successes

p is the probability of success

q is the probability of failure

In this case, n = 14, p = 0.85, and q = 0.15.

(a) To find the probability that all 12 of the flights were on time, we can plug k = 12 into the binomial probability formula:

P(12 successes in 14 trials) = (14 choose 12) * (0.85)^1^2*(0.15)^2

Using a calculator, we can find that this probability is approximately 0.0032.

(b) To find the probability that exactly 10 of the flights were on time, we can plug k = 10 into the binomial probability formula:

P(10 successes in 14 trials) = (14 choose 10) *(0.85)^1^0 *(0.15)^4

Using a calculator, we can find that this probability is approximately 0.1022.

(c) To find the probability that 10 or more of the flights were on time, we can add up the probabilities of 10, 11, 12, 13, and 14 successes:

P(10 or more successes) = P(10 successes) + P(11 successes) + P(12 successes) + P(13 successes) + P(14 successes)

Using a calculator, we can find that this probability is approximately 0.3783.

(d) To determine whether it would be unusual for 11 or more of the flights to be on time, we can find the probability of this event and compare it to a common threshold for unusualness, such as 0.05.

P(11 or more successes) = P(11 successes) + P(12 successes) + P(13 successes) + P(14 successes)

Using a calculator, we can find that this probability is approximately 0.2142. This probability is greater than 0.05, so it would not be considered unusual for 11 or more of the flights to be on time.

Final answer:

This problem can be approached as a binomial distribution. The probability of a particular number of flights on time is calculated using the binomial probability formula. Determining 'unusual' can be subjective but normally a probability less than 0.05 is considered unusual.

Explanation:

This problem is a binomial probability problem because we have a binary circumstance (flight is either on time or it isn't) and a fixed number of trials (14 flights). The binomial probability formula is P(X=k) = C(n, k) * (p^k) * ((1 - p)^(n - k)) where n is the number of trials, k is the number of successful trials, p is the probability of success on a single trial, and C(n, k) represents the number of combinations of n items taken k at a time.

(a) For all 12 flights on time, it seems there's a typo; there are 14 flights in the sample. We can't calculate for 12 out of 14 flights without the rest of the information.

(b) For exactly 10 flights, we use n=14, k=10, p=0.85: P(X=10) = C(14, 10) * (0.85^10) * ((1 - 0.85)^(14 - 10)).

(c) For 10 or more flights on time, it's the sum of the probabilities for 10, 11, 12, 13, and 14 flights on time.

(d) For determining whether 11 or more on-time flights is unusual, it depends on the specific context, but we could consider it unusual if the probability is less than 0.05.

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Solve for the variable 8x + 15 = 63

Answers

Answer:

6

Step-by-step explanation:

63-15 =48 48/8 is  6

Answer:

x = 6

Step-by-step explanation:

8x + 15 = 63

8x = 63 - 15

8x = 48

x = 48/8

x = 6

One number is 3 more than 3 times another. The sum of the numbers is 19. Find the two numbers.The two numbers are
and

Answers

Answer:

number x

y = 3x +3

sum = x + y

19 = x + 3x + 3

19 = 4x + 3

19- 3 = 4x

16 = 4x

16/4 = x

x = 4

BRAINLIEST AND FIVE STARS TO CORRECT ANSWER

Answers

Note that the first function is f(x) = 4 for x less than 2  and for the second function f(x) = -1 for x greater than or equal to 2  

So the limit as x approaches two from the left is 4 and the limit as x approaches 2 from the right is -1.

so looks like you would need the third choice:  4; -1