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-Dominant- [34]
3 years ago
10

5. You sell candy bars for a fundraiser. Each box contains 50 bars and costs $30. If you want to make $45 profit for each box, h

ow much should you sell each bar for?
A. $1.00
B. S1.25
C. $1.50
D. S2.00​
Mathematics
2 answers:
vesna_86 [32]3 years ago
8 0
This makes no sense?
If you sell them a 1.00 then it will be 50 dollars you will collect..
Leokris [45]3 years ago
5 0

Answer: C

Step-by-step explanation:

1.5 x 50 = 75

75 - 30 = 45

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X would be 0.5. Hope this helped
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Which formula can you use to find the radius (r)r if you know that C=2πr?
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Please help i don’t understand
olganol [36]

9514 1404 393

Answer:

  3

Step-by-step explanation:

Vertical angles share a vertex point, and have opposite rays for sides. In short, they are across the point of intersection from each other. Vertical angles do not have a common side (they are <em>not</em> adjacent).

The first part of the problem is to identify angle NCA. That is shown in red in the attachment. The vertical angle is across the point of intersection. It is angle 3.

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3 years ago
-6.3x+14 and 1.5x-6<br> answer in simplified form
koban [17]

Answer:

The simplified form of -6.3x+14 and 1.5x-6 is -4.8x+8

Step-by-step explanation:

We have to simplify the following

-6.3x+14 and 1.5x-6

it can be written as:

=(-6.3x+14) + (1.5x-6)

Adding the like terms

=(-6.3x+1.5x)+(14-6)

= (-4.8x)+(8)

= -4.8x+8

So, the simplified form of -6.3x+14 and 1.5x-6 is -4.8x+8

3 0
4 years ago
Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the
ziro4ka [17]

Answer:

a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}&#10;And in order to find the sample mean we just need to use this formula:&#10;[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0
4 years ago
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