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iris [78.8K]
3 years ago
10

The image of the point (-8, -9) under a translation is (-11, -14). Find the

Mathematics
1 answer:
hjlf3 years ago
5 0

Answer:

N

Step-by-step explanation:

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Find the 13th term of a sequence where the
ValentinkaMS [17]

Answer:

First, you write an explicit formula then plug in the value of n in the formula with the term number you are trying to find, in this case the 13th term.


The steps are in the pictures


the final answer is 121 btw


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3 years ago
What is the total surface area of the rectangular prism?​
bogdanovich [222]

Answer:

Total surface = 102.4 cm^2 (second answer in the list of options)

Step-by-step explanation:

Notice that we need to add the surface of 4 rectangles, two have smaller sizes than the other two. A pair of rectangles are 7.2 cm x 4 cm = 28.8 cm^2, while the other pair has 7.2 cm x 2 cm = 14.4 cm^2

There are also two equal bases each one of size 4 cm x 2 cm = 8 cm^2

Now we add two of each of the individual surfaces discussed above to complete the total surface area:

2 * 28.8 + 2 * 14.4 + 2 * 8 =  102.4 cm^2

which agrees with the second answer option provided.

3 0
3 years ago
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3 0
4 years ago
What is the slope and y intercept of the line y= 4x -12-2x?
aleksandr82 [10.1K]
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6 0
3 years ago
Read 2 more answers
It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha
stira [4]

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.54% of the calls receive a busy signal, hence  p = 0.0054.
  • A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

P(X \geq 5) = 1 - P(X < 5)

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

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

P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009

P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062

P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218

P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513

P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0
2 years ago
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