Which number is equivalent to the fraction 15/7

Sara's savings account balance changed by -$34 one week and by -$67 the next week .Which amount represents the greater change?

fomenos

The greater change is represented by the absolute magnitude

|-34| = 34
|-67| = 67

67 > 34

-67 represents the greater change

4 0

3 years ago

Read 2 more answers

What appears to be the range of the part of the function shown on the grid?

Yuki888 [10]

Answer:

D) g(x) = 2x-8

Step-by-step explanation:

4 0

2 years ago

A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio

yaroslaw [1]

Answer:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

$\hat p=0.45$ estimated proportion of chips that fail in the first 1000 hours of their use

$\mu_0 =0.48$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion si less then 0.48:

Null hypothesis: $p\geq 0.48$

Alternative hypothesis: $p < 0.48$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

6 0

3 years ago

The Copy Shop has made 20 copies of a document for you. Since the defective rate is 0.1, you think there may be some defective c

Pepsi [2]

Answer:

Binomial

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

Step-by-step explanation:

For each copy of the document, there are only two possible outcomes. Either it is defective, or it is not. This means that we can solve this problem using the binomial probability distribution.

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 combinatios 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.

In this problem

Of the 20 copies, 2 are defective, so $p = \frac{2}{20} = 0.1$ .

What is the probability that you will encounter neither of the defective copies among the 10 you examine?

This is P(X = 0) when $n = 10$ .

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

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.3487$

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

8 0

3 years ago

Graph the line with slope -3 passing through the point (5,0)

Ghella [55]

Y-y1= m(x-x1) point-slope form
y-0= -3(x-5)
solve this point-slope form to slope-intercept form. Use the information to graph the line.

4 0

3 years ago