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amm1812
2 years ago
5

Max has a collection of 99 dimes and pennies worth $4.41. How many pennies does he have?

Mathematics
2 answers:
vladimir1956 [14]2 years ago
7 0
Delete or comment answer I'm tired sorry
Natali5045456 [20]2 years ago
4 0

Max has a collection of 99 dimes and pennies worth $4.41.

1 dime = 10 cents

1 penny = 1 cent

1 dollar = 100 cents so 4.41 dollars = 4.41* 100 = 441 cents

Lets 'd' be the number of dimes

Let 'p' be the number of pennies

dimes + pennies = 99

So equation-1 becomes d + p = 99

Worth of dimes and pennies is 440 cents

Worth of 'd' dimes is 10d

Worth of 'p' pennies is 1p

So the equation-2 becomes 10d + 1p = 441

Solve both the equations

10d + 1p = 441

d + p = 99

Multiply the second equation by 10 and then subtract it

10d + 1p = 441

-10d -10p =-990

--------------------

-9p =- 549

Divide by -9 on both sides

So p = 61

Max have 61 pennies .


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Answer:

No, at the 0.05 significance level, the number of units produced on the night shift is not larger.

Step-by-step explanation:

We are given that the mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351.

Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift.

Let \mu_1 = population mean number of units produced on the day shift

      \mu_2 = population mean number of units produced on the night shift

So, <u>Null Hypothesis</u>, H_0 : \mu_1-\mu_2\geq0  or  \mu_1\geq\mu_2    {means that the mean number of units produced on the night shift is same or lesser on the day shift}

<u>Alternate Hypothesis,</u> H_A : \mu_1-\mu_2  or  \mu_1    {means that the mean number of units produced on the night shift is larger}

The test statistics that will be used here is <u>Two-sample z test statistics</u> as we know about population standard deviations;

              T.S.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }  ~ N(0,1)

where, \bar X_1 = sample mean number of units produced by a sample of 54 day-shift workers = 345

        \bar X_2 = sample mean number of units produced by a sample of 60 night-shift workers = 351

       \sigma_1  = population standard deviation of the number of units produced on the day shift = 21

        \sigma_2 = population standard deviation of the number of units produced on the day shift = 28

        n_1 = sample of day-shift workers = 54

        n_2 = sample of night-shift workers = 60

So, <em><u>test statistics</u></em>  =  \frac{(345-351)-(0)}{\sqrt{\frac{21^{2} }{54}+\frac{28^{2} }{60} } }

                              =  -1.302

Now at 0.05 significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -1.302 > -1.6449 so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean number of units produced on the night shift is same or lesser than those produced on the day shift.

4 0
2 years ago
Aster sold 18 oranges. If these are 12% of her total oranges are not sold​
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Answer:

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Step-by-step explanation:

(-15)/3

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Let’s model how to divide the tiles into three equal groups by moving one tile at a time into three separate groups until all 15 tiles have been moved.

Step 1.

Start with 15 red tiles, as in the first picture.

Step 2.

Start by moving tiles one at a time to begin creating three new groups.

The middle picture shows the situation after you have moved the first three tiles.

Step 3.  

Continue moving tiles one at a time from the original group and stacking them evenly in the three new groups until you have no more tiles to move.  

The third picture shows the situation when you have finished.

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Is the answer positive or negative?  The answer is negative, because there are five red negative tiles in each group.

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90 + 150 = 240

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