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DaniilM [7]
3 years ago
7

If 1,529 students participate in field day, how many cases of water bottles should be purchased if the water comes in cases of 1

00?
Mathematics
1 answer:
serious [3.7K]3 years ago
6 0
Dude that’s easy hahahahhahaha
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Greta is trying to determine the portion of green candies in various bags of green and yellow candies. Using the information bel
IrinaK [193]

Answer: a) \dfrac{1}{3} b) \dfrac{71}{100} c) \dfrac{5}{9}

Step-by-step explanation:

Since we have given that

There are green and yellow candies in each bag.

Bag A: Two thirds of the candies are yellow. What portion of the candies is green?

Part of yellow candies in bag A = \dfrac{2}{3}

Part of green candies in bag A would be

1-\dfrac{2}{3}\\\\=\dfrac{3-2}{3}\\\\=\dfrac{1}{3}

Bag B: 29 % of the candies are yellow. What portion of the candies is green?

Percentage of candies are yellow = 29%

Portion of candies are green is given by

1-\dfrac{29}{100}\\\\=1-0.29\\\\=0.71\\\\=\dfrac{71}{100}

Bag C: 4 out of every 9 candies are yellow. What portion of the candies is green?

Portion of yellow candies = \dfrac{4}{9}

Portion of green candies would be

1-\dfrac{4}{9}\\\\=\dfrac{9-4}{9}\\\\=\dfrac{5}{9}

Hence, a) \dfrac{1}{3} b) \dfrac{71}{100} c) \dfrac{5}{9}

6 0
3 years ago
PLSSS HELPPPPP MEEEE III WILLL GIVVVEEE BRAINLIESTTTT!!!!!!
Artyom0805 [142]
They are equal, 8 x 7 = 56 and 7 x 7 = 49
6 0
3 years ago
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

6 0
3 years ago
Someone pls help asap this is supposed to be done it 3 min
insens350 [35]

Step-by-step explanation:

$75      30%      0.70(75) = x      $52.50

$18       65%      0.35(18) = x      $6.30

$60      30%     0.70x = 42        $42

$35       20%     0.80x = 28       $28

$150      25%     0.75(150) = x     $112.50

6 0
2 years ago
Read 2 more answers
James had $15,345.21 withheld from his income one year. He
puteri [66]

Answer:

$2,373.23

Step-by-step explanation:

-Tax refund is determined by comparing your total income tax to the amount that was withheld for federal income tax.

-Assuming that the amount withheld for federal income tax was greater than your income tax for the year, you will receive a refund for the difference.

#Refund =Withheld -Actual Tax liability=$15345.21-$12971.98=$2,373.23

Hence James' tax refund will be $2,373.23

5 0
3 years ago
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