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MAXImum [283]
3 years ago
11

a local charity held a crafts fair selling donated,handmade items. Total proceeds from the sale were $1,875. A total of 95 items

were sold,some at $15 each and the rest at $25 each. Let x be the number of $15 items and y the number of $25 items. How many items sold at $25?
Mathematics
1 answer:
Lana71 [14]3 years ago
5 0
15x+25y=1875
Let
y=95-x
15x+25 (95-x)=1875
Solve for x
15x+2375-25x=1875
15x-25x=1875-2375
-10x=-500
X=500/10
X=50 the number of $15 items

Y=95-50=45 the number of $25 items
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Y=-3x + 4
serg [7]
Use substitution
Set them equal to each other
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Solve the equation
4 = 6x - 2
6x = 6, x = 1
Plug in 1 for any equation to find y
Y = -3(1) + 4
Y = -3 + 4, y = 1
Solution: x = 1, y = 1
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Solve for r.<br> r+2&lt;10 <br> FIRST TO SOLVE WILL BE BRAINLIEST
Marat540 [252]

Answer:

The answer is:

r<8

Alternative Form of the answer:

r ∈ (-∞, 8)

4 0
3 years ago
A sample of eight workers in a clothing manufacturing company gave the following figures for the amount of time(in minutes) need
lilavasa [31]

Answer:

10.108 < \mu < 13.892    

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

We have the following distribution for the random variable:

X \sim N(\mu , \sigma=0.45)

And by the central theorem we know that the distribution for the sample mean is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

2) Confidence interval

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=12

The sample deviation calculated s=2.268

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=8-1=7

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that t_{\alpha/2}=2.36

Now we have everything in order to replace into formula (1):

12-2.36\frac{2.268}{\sqrt{8}}=10.108    

12+2.36\frac{2.268}{\sqrt{8}}=13.892

So on this case the 95% confidence interval would be given by (10.108;13.892)

10.108 < \mu < 13.892    

7 0
3 years ago
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