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2 years ago

Find the product of 0.094 ×0.367

Mathematics

1 answer:

Keith_Richards [23]2 years ago

3 0

0.034498 is the answer

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You jog 6/23 miles around a track each day. If you jogged that distance 4 times last week, how many miles did you jog?

LenaWriter [7]

Well, since you jogged 6/23 mi. a day, for 4 days, it'll be (6/23)×4. This is 24/23 which is one mile and 1/24 of one.

6 0

2 years ago

A Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer; a sample

8090 [49]

Answer:

Step-by-step explanation:

Step(i):-

Given first random sample size n₁ = 500

Given Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer.

First sample proportion

$p^{-} _{1} = \frac{65}{500} = 0.13$

Given second sample size n₂ = 700

Given a sample of 700 men (unrelated to the women) ages 18-29 found that 133 men said that they had the most say when purchasing a computer.

second sample proportion

$p^{-} _{2} = \frac{133}{700} = 0.19$

Level of significance = α = 0.05

critical value = 1.96

Step(ii):-

Null hypothesis : H₀: There is no significance difference between these proportions

Alternative Hypothesis :H₁: There is significance difference between these proportions

Test statistic

$Z = \frac{p_{1} ^{-}-p^{-} _{2} }{\sqrt{PQ(\frac{1}{n_{1} } +\frac{1}{n_{2} } )} }$

where

$P = \frac{n_{1} p^{-} _{1}+n_{2} p^{-} _{2} }{n_{1}+ n_{2} } = \frac{500 X 0.13+700 X0.19 }{500 + 700 } = 0.165$

Q = 1 - P = 1 - 0.165 = 0.835

$Z = \frac{0.13-0.19 }{\sqrt{0.165 X0.835(\frac{1}{500 } +\frac{1}{700 } )} }$

Z = -2.76

|Z| = |-2.76| = 2.76 > 1.96 at 0.05 level of significance

Null hypothesis is rejected at 0.05 level of significance

Alternative hypothesis is accepted at 0.05 level of significance

Conclusion:-

There is there is a difference between these proportions at α = 0.05

3 0

2 years ago

Consider a single spin of the spinner.

alex41 [277]

Answer:

"landing on a shaded portion and landing on a 3"

"landing on an unshaded portion and landing on a number less than 2 "

Step-by-step explanation:

Mutually exclusive means the events will have no intersection.

Let's look at your first choice:

"landing on a shaded portion and landing on an even number"

Landing on a shaded portion would be 1 or 4.

Landing on an even number would be 2 or 4.

There is an intersection (they contain a common element), the 4.

These events are not mutually exclusive.

Let's look at your second choice:

"landing on a shaded portion and landing on a number greater than 3"

Landing on a shaded portion would be 1 or 4.

Landing on a number greater than 3 would be just 4.

There is an intersection; they both contain 4.

These events are not mutually exclusive.

Let's look at your third choice:

"landing on a shaded portion and landing on a 3"

Landing on a shaded portion would be 1 or 4.

Landing on 3 would just be 3.

There is no common elements in the lists listed. These events have no intersection.

These events are mutually exclusive.

Let's look at your fourth choice:

"landing on an unshaded portion and landing on an odd number"

Landing on a unshaded portion would be 2 or 3.

Landing on an odd number would be 1 or 3.

There is an intersection; they both have 3 in common.

These events are not mutually exclusive.

Let's look at your fifth choice:

"landing on an unshaded portion and landing on a number less than 2 "

Landing on an unshaded portion would 2 or 3.

Landing on a number less than 2 would be 1.

There is no intersection.

These events are mutually exclusive.

6 0

2 years ago

Read 2 more answers

According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me

Reil [10]

Answer:

a) $\bar X=10.65$

$Median =\frac{10.7+10.7}{2}=10.7$

$Mode= 10.7$

b) $Range = 11.8-8.3=3.5$

$s= 0.948$

c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

$s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0

2 years ago

Employees in the marketing department of a large regional restaurant chain are researching the amount of money that households i

kap26 [50]

Answer:

The z-score for the group "25 to 34" is 0.37 and the z-score for the group "45 to 54" is 0.25.

Step-by-step explanation:

The data provided is as follows:

25 to 34 45 to 54

1329 2268

1906 1965

2426 1149

1826 1591

1239 1682

1514 1851

1937 1367

1454 2158

Compute the mean and standard deviation for the group "25 to 34" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [1329+1906+...+1454]=\frac{13631}{8}=1703.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1086710.875}=394.01$

Compute the z-score for the group "25 to 34" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1703.875}{394.01}=0.3734\approx 0.37$

Compute the mean and standard deviation for the group "45 to 54" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [2268+1965+...+2158]=\frac{14031}{8}=1753.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1028888.875}=383.39$

Compute the z-score for the group "45 to 54" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1753.875}{383.39}=0.25333\approx 0.25$

Thus, the z-score for the group "25 to 34" is 0.37 and the z-score for the group "45 to 54" is 0.25.

8 0

2 years ago