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Artist 52 [7]
3 years ago
11

at the local deli chrystelle bought 2/3 pound of ham and 3/5 as many pounds of cheese as ham. cheese costs 4.95 per pound and ha

m costs 5/3 as much per pound as cheesem which statements are correct? A. chrystelle will pay 2.97 per pound for the ham. B. she asked for 2/5 pound of cheese. C. she will pay a total of 5.50 for the ham. D. she will pay a total of 8.25 for the cheese E. she will pay 7.48 combined for the ham and the cheese.
Mathematics
1 answer:
Zielflug [23.3K]3 years ago
8 0
Chrystelle bought 2/3 pound of ham. (given)
She bought 3/5 * 2/3 = 2/5 pound of cheese.

The cheese cost 4.95 per pound, so she paid 2/5*4.95 = 1.98 for the cheese.
The ham cost 5/3 * 4.95 = 8.25 per pound, so she paid 2/3*8.25 = 5.50 for the ham.

Her total purchase was 1.98 +5.50 = 7.48 for the ham and cheese.

Selections B, C, E are correct.
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Find two consecutive numbers whose sum is 117
Serhud [2]
So,

x + y = 117
y = x + 1
x + x + 1 = 117

Collect Like Terms
2x + 1 = 117

Subtract 1 from both sides
2x = 116

Divide both sides by 2
x = 58

y = x + 1

y = 58 + 1

y = 59

The 2 consecutive integers are 58 and 59.
6 0
3 years ago
Read 2 more answers
A box contains 10 transistors, 4 of which are defective. If 3 are sold at random, find the probability that exactly 1 are defect
Readme [11.4K]
I'd say around 10%
4 out of 10 = 40%
3 out of the 10 are sold = 30%
1 out of the 3 might be bad = 10%
I may be wrong.
3 0
3 years ago
Plot the following points and find the area of the figure: (3,5); (-1,5); (-1,-3); (3,-3)
Westkost [7]

Answer:

im sorry i cant help you without a picture

Step-by-step explanation:

5 0
3 years ago
A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting reading
valkas [14]

Answer:

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

p_v =2*P(t_{11}  

Step-by-step explanation:

1) Data given and notation  

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}

The results obtained are:

\bar X=98.375 represent the sample mean  

s=0.6.109 represent the sample standard deviation  

n=12 sample size  

\mu_o =100 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :  

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

4) P-value  

First we need to find the degrees of freedom for the statistic given by:

df=n-1=12-1=11

Since is a two sided test the p value would given by:  

p_v =2*P(t_{11}  

5) Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.  

3 0
3 years ago
X-5+x-11=100 what is the reason for this statement for geometry?​
Mariulka [41]

Answer:

because when the value of x is replaced in the factorisation it Will be equivalent to 100

3 0
2 years ago
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