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Verizon [17]
3 years ago
10

Wendy needs to purchase 25 vases, which cost $3 each, and flowers for the vases, which cost $2 each. She has $275 to spend on he

r vases and flowers. Which of the following inequalities would show the maximum number of flowers, x, Wendy can buy without spending more than $275?
A. $2x + $75 < $275
B. $2x + $75 > $275
C. x + $25 > $275
D. x + $25 < $275
Show all work!
Mathematics
1 answer:
Levart [38]3 years ago
5 0
The answer is A.

The cost of 25 vases would be $75 (25*3). then x would be the number of flowers she can buy at $2 each before she goes over $275. So, $2x+$75 < $275
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The point (8,k) lies on the line y=3x-5
Contact [7]

Answer:

x times 5 is 21

Step-by-step explanation:

you times x to 5 and then you will get 21

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3 years ago
A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds tha
ollegr [7]

Answer:

99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Step-by-step explanation:

We are given that a high school principal wishes to estimate how well his students are doing in math.

Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

                          P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of students received a passing grade = 77%

           n = sample of tests = 40

           p = population proportion

<em>Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                           level of significance are -2.5758 & 2.5758}  

P(-2.5758 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<u>99% confidence interval for p</u> = [\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

 = [ 0.77-2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } } , 0.77+2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } } ]

 = [0.5986 , 0.9414]

Therefore, 99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Lower bound of interval = 0.5986

Upper bound of interval = 0.9414

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Find the solution to the systems of equations using the elimination method.
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9w = -54
Virty [35]
0 - (-6)


<span>-4 - 5

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What is 2.5 simplified
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Step-by-step explanation:

2.5 =  \frac{25}{10}  =  \frac{5 \times 5}{2 \times 5}  =  \frac{5}{2}  \\

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