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Mice21 [21]
3 years ago
14

I need help now please

Mathematics
2 answers:
aliina [53]3 years ago
7 0

Answer:

4

Step-by-step explanation:

7 4

3 × (-3)

It would be answer four because it would be equivalent and equal expression

Hope that helps :)

Juliette [100K]3 years ago
5 0
Answer:
Correct option: A. 3^7x1/3^4

Hope it helps u
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Solve for t.
iVinArrow [24]

Answer:

4t - 3 > 37t - 50

Step-by-step explanation:

Step 1: 4t - 3 > 37t - 50

          <u>-4t         -4t</u>

           -3 > 33t - 50

          <u>+50        + 50</u>

        <u>47</u> > <u>33</u>t

        33    33

        <em>t < 47/33</em>

3 0
2 years ago
All of the points in the graph are on the same line. Find the slope of the line.
Klio2033 [76]

Answer:

Step-by-step explanation:

take (1,3) and (9,7)

slope=(7-3)/(9-1)=4/8=1/2

3 0
3 years ago
Solve by factoring 7b^2-8b+3=2
Ahat [919]
7b^2-8b+3=2 \\ 7b^2-8b+1=0 \\ (7x-1)(x-1)=0 \\ x=1 \\ x= \frac{1}{7}
6 0
3 years ago
Read 2 more answers
(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice
vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

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.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that \mu = 100, \sigma = 15

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{107.5 - 100}{15}

Z = 0.5

Z = 0.5 has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{130 - 100}{15}

Z = 2

Z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549

P(X \geq 1) = 1 - P(X = 0) = 0.0451

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

8 0
2 years ago
The useful life of a radial tire is normally distributed with a mean of 30,000 miles and a standard deviation of 5000 miles. The
otez555 [7]

Answer:

  empirical rule: 81.5%

  table or calculator: 81.9%

Step-by-step explanation:

The lower limit of the life range of interest has a z-score of ...

  z = (x -μ)/σ = (20,000 -30,000)/5,000 = -2

The upper limit has a z-score of ...

  z = (35,000 -30,000)/5,000 = 1

<u>Empirical rule solution</u>

The empirical rule tells you that 95% of the distribution lies within 2 standard deviations of the mean, so (100% -95%)/2 = 2.5% lie below z = -2. It also tells you 68% lie within 1 standard deviation of the mean, so (100% -68%)/2 = 16% lie above z = 1.

The fraction that lies within -2 to 1 standard deviations of the means is thus ...

  (100% -2.5% -16%) = 81.5%

The probability the tire has a life in the desired range is about 81.5%.

__

<u>Calculator solution</u>

A probability calculator for the Normal distribution tells you that ...

  P(-2 < z < 1) ≈ 0.8185946

The probability the tire has a life in the desired range is about 81.9%.

6 0
2 years ago
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