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Seven students chose integers: -16, 12, -13, -6, 5, 6 and 1 Order the numbers from least to greatest. When all numbers are order

Lorico [155]

Answer:-16<-13<-6<1<5<6<12

1

-16

12

Step-by-step explanation:

8 0

3 years ago

Kyle has 22 more cards than 3 times the number of cards that Stan owns. Together, they own less than 58 cards. How many cards ar

Slav-nsk [51]

If its only one answer its A

5 0

3 years ago

Read 2 more answers

1-2m-5m=15 multistep homework

Montano1993 [528]

1-2m-5m=15

combine like terms

1 -7m = 15

subtract 1 from each side

1-1-7m = 15-1

simplify

-7m = 14

divide by -7

-7m/-7 = 14/-7

simplify

m = -2

6 0

3 years ago

What’s the class width

aksik [14]

Answer:

4

Step-by-step explanation:

Class width is said to be the difference between the upper class limit and the lower class limit consecutive classes of a grouped data. To calculate class width, this formula can be used:

CW = UCL - LCL

Where,

CW= Class width

UCL= Upper class limit

LCL= Lower class limit

From the table above:

For class 1, CW = 64 - 60 = 4

For class 2, CW = 69 - 65 = 4

For class 3, CW = 74 - 70 = 4

For class 4, CW = 79 - 75 = 4

For class 5, CW = 84 - 80 = 4

Therefore, the class width of the grouped data = 4

6 0

3 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

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