Enter the equivalent distance in km in the box.

In a recent survey 55% of pet owners have more than one pet if there was 620 pet owners survey which proportion could be used to

Alika [10]

Answer:

341..

Step-by-step explanation:

i would personally say 341 because 620 multiplied by 55% or .55, as written on a calculator, would be 341.

5 0

3 years ago

Which fraction below represents a repeating decimal?

nikklg [1K]

Answer:

22/12

Step-by-step explanation:

3 0

1 year ago

The formula F=95C+32 changes a temperature reading from the Celsius scale C to the Fahrenheit scale F. What is the temperature m

Law Incorporation [45]

Answer:

122°F

Step-by-step explanation:

F=(50×9/5)+32

= 122°F

8 0

2 years ago

In recent commercials Toyota claimed that 80% of its vehicles sold in the past 20 years are still on the road. You work for an a

Anvisha [2.4K]

Answer:

Step-by-step explanation:

We have to perform one sample proportion test. To do so we have to proceed through following steps.

(1) We have to collect data about vehicle sales in last 20 years.

(2) Using random number table or random number generator or otherwise we have to select sample of vehicles from the list of sale. In this step, we have to keep an eye on the fact that sample is to be selected from all years (all of 20). Using stratified random sampling by dividing vehicles sold over different years is an effective way to do so. That means, we may divide vehicles sold into 20 homogeneous strata over 20 years and select sample from each strata.

(3) After selecting sample of vehicles we have to communicate to corresponding vehicle owners to gather information about that particular vehicle.

(4) We have to note number of vehicles which are still on the road.

Suppose, we have selected n vehicles as sample and after communicating we found that \tiny r of those are still on the road.

We have to test for null hypothesis $H_0:p=0.80$

against the alternative hypothesis $H_1 \neq 0.80$

Our test statistic is given

$z=\frac{\hat p - p_0}{\sqrt{p_0(1-p_0)/n} }$

here,

$\hat p =\frac{r}{n}$

$p_0 =0.80$

$z=\frac{r/n-0.80\sqrt{n} }{0.40} \\\\=2.5(\frac{r}{n} -0.8)\sqrt{n}$

We then calculate corresponding p-value.

We reject our null hypothesis if $\text{p-value}$ , level of significance.

According to our obtained p-value and level of significance we proceed to certain conclusion.

8 0

3 years ago

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil

Fiesta28 [93]

Answer:

The sample has not met the required specification.

Step-by-step explanation:

As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow

$H_0$ : The true average penetration is 50 mils

$H_a$ : The true average penetration is > 50 mils

Since we are trying to see if the true average is greater than 50, this is a right-tailed test.

If the level of confidence is α = 0.05 then the $z_\alpha$ score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)