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What is the constant of variation k of the line y=kx through (5,8)

mina [271]

Answer:

k= 8/5

Step-by-step explanation:

here x= 5 and y= 8

we have y=kx

k= y/x

k= 8/5 this is constant of variation

8 0

3 years ago

Read 2 more answers

If shape 1 is similar to shape Il and shape II is similar to shape III, what does that mean for shape I and shape III?

blagie [28]

Answer:

Step-by-step explanation:

They are also similar.

Both have 4 right angles. Both have 2 sets of parallel lines. Both have 4 equal sides.

But I think you are supposed to answer if figure 1 is similar to figure 2 and figure two is similar to figure 3 then by the transitive property, figure 1 is similar to figure 3.

7 0

3 years ago

Antibiotics in Infancy

leonid [27]

Answer:

e)The information from the sample does not give enough information to support that more than 70% of Canadian children receive antibiotics during the first year of life

Step-by-step explanation:

The proportion we will use for the test is p = 70 %

From a random sample we got

n = 616

x = 438 then p₁ = 438/616 p₁ = 0,71 p₁ = 71 %

Then q₁ = 1 - p₁ q₁ = 1 - 0,71 q₁ = 0,29 q₁ = 29 %

a) Hypothesis test:

Null hypothesis H₀ p₁ = p

Alternative hypothesis Hₐ p₁ > p >70 %

CI = 95 % significance level α = 5 % α = 0,05

z(c) for α = 0,05 from z-table is z(c) = 1,64

b) To calculate z(s) = ( p₁ - p ) / √ p₁*q₁ / n

z(s) = ( 0,71 - 0,70 )/ √ 0,71*0,29/616

z(s) = 0,01 /0,01828

z(s) = 0,547 ≈ 0,55

c) p-value for z(s) is from z-table p-value ≈ 0,7088

d) p-value > 0,05 then we accept H₀ we don´t have enough evidence to reject H₀

e)The information from the sample does not give enough information to support that more than 70% of Canadian children receive antibiotics during the first year of life

3 0

3 years ago

A TV station claims that 38% of the 6:00 - 7:00 pm viewing audience watches its evening news program. A consumer group believes

Elena L [17]

Answer:

$z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374$

$p_v =P(Z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ 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 people that regularly watch the TV station’s news program is significantly less than 0.38 .

Step-by-step explanation:

1) Data given and notation

n=830 represent the random sample taken

X=282 represent the people that regularly watch the TV station’s news program

$\hat p=\frac{282}{830}=0.340$ estimated proportion of people that regularly watch the TV station’s news program

$p_o=0.38$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.38:

Null hypothesis: $p\geq 0.38$

Alternative hypothesis: $p < 0.38$

When we conduct a proportion test we need to use the z statisitc, 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 $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374$

4) 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 very low value and using the significance level given $\alpha=0.05$ we have $p_v$ 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 people that regularly watch the TV station’s news program is significantly less than 0.38 .

3 0

3 years ago

What is the answer to this question?

Pie

Answer:

-9

Step-by-step explanation:

Add the cards together

2 + -6 + 1 + -2 + -4

-9

4 0

3 years ago