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alex41 [277]
2 years ago
8

The cost of 6 sandwiches and 4 drinks is $53. The cost of 4 sandwiches and 6 drinks is $47 how much does one sandwich cost

Mathematics
2 answers:
siniylev [52]2 years ago
6 0
$53-$47=$6
2 sandwiches are $6
so 1 sandwich is $3
devlian [24]2 years ago
3 0
Let sandwiches and drinks be s and d resp so according to the question,
by 1st case,
6s + 4d = 53
or, d = (53-6s)/4
again by 2nd case,
4s + 6 d = 47
4s + 6(53-6s)/4 = 47
or, (8s +159-18s)/2=47
or 159-10s = 94
or 65 =10s
or, s = $6.5
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part a: answer b. 0.75x + 0.15y = 90

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Which of the following is the missing side length that completes the Pythagorean triple below? 5, 12, ____ A. 17 B. 15 C. 13 D.
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Round to the nearest benchmark fraction 5/9
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A random sample from a normal population is obtained, and the data are given below. Find a 90% confidence interval for . 114 157
melamori03 [73]

Answer:

103.160 \leq \sigma \leq 187.476

And the upper bound rounded to the nearest integer would be 187.

Step-by-step explanation:

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 mean or variance 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.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

\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  

Solution to the problem

Data given: 114 157 203 257 284 299 305 344 378 410 421 450 478 480 512 533 545

The confidence interval for the population variance \sigma^2 is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \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}^{17} (x_i -\bar x)^2}{n-1}}

And in order to find the sample mean we just need to use this formula:

\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample mean obtained on this case is \bar x= 362.941 and the deviation s=132.250

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

df=n-1=17-1=16

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a tabel to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,16)" "=CHISQ.INV(0.95,16)". so for this case the critical values are:

\chi^2_{\alpha/2}=26.296

\chi^2_{1- \alpha/2}=7.962

And replacing into the formula for the interval we got:

\frac{(16)(132.250)^2}{26.296} \leq \sigma \leq \frac{(16)(132.250)^2}{7.962}

10641.959 \leq \sigma^2 \leq 35147.074

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

103.160 \leq \sigma \leq 187.476

And the upper bound rounded to the nearest integer would be 187.

4 0
2 years ago
PLEASE HELP ME!
dezoksy [38]
Thicknesses at different point are: <span>41, 38, 36, 29, 34, 44, 46, 43, 35, 40


In increasing order: 29, 34, 35, 36, 38, 40, 41, 43, 44, 46

Median = (38+40)/2 = 39m</span>

Median thickness is 39m
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2 years ago
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