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3 years ago

Answer the question below from I ready

Mathematics

2 answers:

nika2105 [10]3 years ago

7 0

The answer is #1 hope this helps

san4es73 [151]3 years ago

6 0

Your answer is going to be the first one: 3^2

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Write a system of equations using x and y in each for the following

vlada-n [284]

1) x+y = 170
x-y = 26

2) Subtract the two equations from each other. Doing this you get 2y = 56.
So the value of y is 28. Then plug in the value of y into any of the two equations and solve for x.

x-28 = 26
x = 54

3) When writing solutions as an ordered pair, the x value always comes first followed by the y value. So it would be (54, 28).

4 0

3 years ago

A manufacturer of shipping boxes has a box

MissTica

then ship that f...... box somewhere becuase all i can think of unless your trying to steal it.

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3 years ago

Consider the probability that greater than 26 out of 124 software users will call technical support. Assume the probability that

Mekhanik [1.2K]

Answer:

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

100% probability that greater than 26 out of 124 software users will call technical support.

Step-by-step explanation:

Test if the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

It is needed that:

$np \geq 10$ and $n(1-p) \geq 10$

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.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Out of 124 software users

This means that $n = 124$

Assume the probability that a given software user will call technical support is 97%.

This means that $p = 0.97$

Conditions:

$np = 124*0.97 = 120.28 \geq 10$

$n(1-p) = 124*0.03 = 3.72 < 10$

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

Consider the probability that greater than 26 out of 124 software users will call technical support.

The lowest possible probability of those is 27, so, if it is 0, since it is considerably below the mean, 100% probability of being greater. We have that:

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

$P(X = 27) = C_{124,27}.(0.97)^{27}.(0.03)^{97} = 0$

1 - 0 = 1

100% probability that greater than 26 out of 124 software users will call technical support.

3 0

3 years ago

The Eastons traveled 190 miles on a trip to Grandma’s house. At the end of the trip the odometer read 7,602 miles. Write and sol

Black_prince [1.1K]

Answer:

X+190=7602

X=7412

Step-by-step explanation:

The odometer read 7412 miles at the beginning of the trip.

3 0

3 years ago

Help quick pleaseeee

ehidna [41]

Answer:

x= 3.096

Step-by-step explanation:

divide both sides by 5: 3^(x-1)= 10

then take the log to get log_3 (10) = x-1 = 2.096

x= 3.096

4 0

2 years ago