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skad [1K]
3 years ago
5

2. Round off your answer to the nearest tenth: 2(-11)-5(-6) 7–3(-2)

Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0

Answer:

12

Step-by-step explanation:

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What is the GCF of 100 and 20
alexdok [17]

Answer:

20

Step-by-step explanation:

20 is the GCF or greatest common factor of 20 and 100.

4 0
2 years ago
At what point on the curve y = 19 2x is the tangent line perpendicular to the line 20x 4y = 1?
Gennadij [26K]

The points are; (7/2, -1/2).

<h3>What is the given point?</h3>

Now the tangent line is given as;20x 4y = 1. When we rewrite it in the slope intercept form, we have the equation as; y = 1 - 20x/4 or y = 1/4 - 5x.

Then to obtain the slope of the curve we have; y = 19 - 2x

dy/dx = 2

Using the relation;

m1m2 = -1

m2 = -1/2

Hence;

y = 1/4 - 5(-1/2)

y = 1/4 + 5/2

y = 7/2

Thus the points are; (7/2, -1/2)

Learn more about normal curve:brainly.com/question/10664419

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7 0
2 years ago
7. Agrupa para sumar.<br>-5 +8 +(-2)​
Brums [2.3K]

Answer:

Obtendrías 1

Step-by-step explanation:

Primero resta 5 de 8 para obtener 3,

luego reste por 2 para obtener 1

¡Espero eso ayude!

6 0
3 years ago
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3 0
3 years ago
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(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
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