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2 years ago

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If 1/6 of the books in the library are cookbooks and 2/5 of the cookbooks are vegetarian cookbooks what fraction of the books in

the library are vegetarian cookbooks
A. 1/30 B. 1/15 C. 1/2 D. 3/10

2 answers:

Serga [27]2 years ago

8 0

Answer:

A. 1/30 of the cookbook

STALIN [3.7K]2 years ago

6 0

Answer:

1/15

Step-by-step explanation:

Took the quiz

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The production department has installed a new spray machine to paint automobile doors. As is common with most spray guns, unsigh

Nesterboy [21]

Answer:

The numbers of doors that will have no blemishes will be about 6065 doors

Step-by-step explanation:

Let the number of counts by the worker of each blemishes on the door be (X)

The distribution of blemishes followed the Poisson distribution with parameter $\lambda =0.5$ / door

The probability mass function on of a poisson distribution Is:

$P(X=x) = \dfrac{e^{- \lambda } \lambda ^x}{x!}$

$P(X=x) = \dfrac{e^{- \ 0.5 }( 0.5)^ x}{x!}$

The probability that no blemishes occur is :

$P(X=0) = \dfrac{e^{- \ 0.5 }( 0.5)^ 0}{0!}$

$P(X=0) = 0.60653$

P(X=0) = 0.6065

Assume the number of paints on the door by q = 10000

Hence; the number of doors that have no blemishes is = qp

=10,000(0.6065)

= 6065

3 0

2 years ago

In ΔVWX, x = 5.3 inches, w = 7.3 inches and ∠W=37°. Find all possible values of ∠X, to the nearest 10th of a degree.

STALIN [3.7K]

Answer:

que?

Step-by-step explanation:

4 0

1 year ago

Read 2 more answers

Glenna wants to rent a car for a trip to key west for one weekshe calls two car rental companies to get prices. mr kotters renta

Korvikt [17]

Answer: Barbarino's rentals has a better deal.

She has to drive 887.5 miles to spend the same amount at either company.

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

<em>Mr.kotters rentals (A) </em>

<em>$99 PER WEEK </em>
<em>$0.11per mile over 100 miles </em>

<em>Barbarino's rentals (B) </em>

<em>$75 per week </em>
<em>$0.15 per mile over 150 miles </em>

For "A"

Cost = 0.11 (432-100) + 99 = $135.52

For "B"

Cost= 0.15 (432-150) +75 = $117.3

Barbarino's rentals has a better deal, since $117.3(B) < $135.52 (A)

To find how many miles would Glenna drive before she would be spending the same amount at either company:

A =B

0.11 (M-100) + 99 =0.15 (M-150) +75 = $117.3

Solving for M (miles)

0.11 M -11+99 = 0.15 M -22.5+75

-11 +99 +22.5 -75 =0.15M -0.11 M

35.5 = 0.04M

35.5/0.04 = M

887.5 =M

She has to drive 887.5 miles to spend the same amount at either company.

4 0

1 year ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

1 year ago

Let ff be the function given by f(x)=e2x−1 x f(x)=e2x−1 x. Which of the following equations expresses the property that f(x)f(x)

navik [9.2K]

Answer:

Step-by-step explanation:

$\lim_{x \to \ 0} \frac{e^{2x}-1 }{x} \\= \lim_{x \to \0} \frac{e^{2x}-1 }{2 x} *2\\\rightarrow2 log e\\\rightarrow 2$

5 0

2 years ago

Read 2 more answers