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2 years ago

11

Diana sold 336 muffins at the bake sale.Bob sold 287 muffins.Bob estimates that he sold 50 fewer muffins than Diana.how did he e

stimate? Explain.

2 answers:

Jet001 [13]2 years ago

7 0

He rounded 336 to 340 and 287 to 290. Then he subtracted 290 from 340, that's how he got 50.

malfutka [58]2 years ago

3 0

Step-by-step explanation:

Since we have given that

Number of muffins Diana sold at the bake sale = 336

Number of muffins Bob sold at the bake sale = 287

Now, here we are using the method of estimation i.e. rounding the integers to nearest ones.

By estimation, we get

Number of muffins Diana sold at the bake sale = 340

Number of muffins Bob sold at the bake sale = 290

So,

Difference between them is given by

$340-290\\=50$

Hence, Bob sold 50 fewer muffins than Diana.

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If the average (arithmetic mean ) of 2,7, and x is 12, what is the value of x?

masya89 [10]

Answer:

x=27

Step-by-step explanation:

The mean is add all the numbers and divide by the number of points

(2+7+x)/3 =12

Multiply each side by 3

(2+7+x)/3 *3 =12*3

2+7+x = 36

Combine like terms

9+x = 36

Subtract 9 from each side

9+x-9 = 36-9

x = 27

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James is the manager at an entertainment arena that draws an average 7,000 patrons per event. Each ticket taker can process 350

Elina [12.6K]

Answer:

No. James didn't have enough ticket takers to process the average number of patrons hat usually attend the events.

He need to hire 2 more ticket taker (i.e 20 ticket takers) in order to process the average number of patrons hat usually attend the events.

Step-by-step explanation:

Given:

Average amount of patron per event= 7000

Each ticket taker can process = 350

Number of ticket takers hired = 18

We need to find the whether he hire enough to process the average number of patrons that usually attend the events.

Solution:

We will find the Average amount of patron ticket takers can process.

Now we can say that

Average amount of patron ticket takers can process is equal to Each ticket taker can process multiplied by Number of ticket takers hired.

framing in equation form we get;

Average amount of patron ticket takers can process = $350 \times 18 = 6300\ patrons$

Hence With the required hires James cannot fully process the patrons usually attending the event.

To find how many ticket takers required we will divide average number of patrons with Each ticket taker can process.

framing in equation form we get;

ticket takers required = $\frac{7000}{350}=20$

hence In order t process the the average number of patrons that usually attend the events James will require 20 ticket takers.

6 0

2 years ago

The cost for a different taxi company is expressed with the equation y = 1.65x + 2.35, where x represents the miles driven and y

Rudiy27

To solve this, all you have to do is plug in the 14 where the x value is:
y=1.65(14) +2.35 which equals 25.45

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The charts show the effect of an interest rate adjustment on an adjustable-rate mortgage.

Tcecarenko [31]

Answer:

C. a higher monthly payment.

Step-by-step explanation:

<h3>Answer options:</h3>

Rate adjustment resulted in

<u>A. a lower interest rate. </u>

No, interest rate increased from 0.4 to 0.5 percent

<u>B. a higher principal. </u>

No, principal decreased from $200000 to $187000

<u>C. a higher monthly payment. </u>

Yes, monthly payment increased from $1059.85 to $1162.56

<u>D. a higher number of payments.</u>

No, number of payments remained as 336

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2 years 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

2 years ago