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Ksivusya [100]
3 years ago
8

If 4 tickets to a show cost $9.00, find the cost of 18 tickets.

Mathematics
1 answer:
NeTakaya3 years ago
3 0
$40.50 you divide 18/4 to see how many times to multiply it then it’s 4.5. Multiply 4.5 by 4 and it’s $40.50
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Weekly demand for a particular item averages 30 units, with a standard deviation of 4. This item is managed with a fixed-order-i
bezimeni [28]

Answer:  93 units should be ordered.

Step-by-step explanation:

Since we have given that

Standard deviation = 4

Average = 30 units

Number of weeks in order interval = 3

Desired service level = 97.5%

Number of units on hand = n = 43

So, the expected  value would be

E[x]=30\times (3+1)+1.96\times 4\sqrt{3+1}-43\\\\E[x]=30\times 4+1.96\times 4\times 2-43\\\\E[x]=120+15.68-43\\\\E[x]=92.68\approx 93

Hence, 93 units should be ordered.

5 0
4 years ago
To prepare 450 ml of a 30% (w/v) solution of potassium chloride (kcl), how many grams of solute are needed?
tiny-mole [99]
The important clue from the problem, would be (w/v). This would indicate that the percent has units of weight per volume. We calculate as follows:

Grams solute = .30 g solute / mL solution (450 mL) = 135 g of solute. 

Hope this answers the question. Have a nice day.
4 0
4 years ago
X+4.9=7.28 x= what is the answer
Mandarinka [93]

Answer:

2.38

Step-by-step explanation:

subtract 4.9 on both sides

8 0
4 years ago
Read 2 more answers
Unit 4 linear equations homework 3 answer sheet
Wittaler [7]

Answer:

utrtyutruytr

Step-by-step explanation:

8 0
4 years ago
1. The table shows the probabilities of a response chocolate or vanilla when asking a child or adult. Use the formula for condit
Stels [109]
\begin{matrix}&\text{chocolate}&\text{vanilla}&\text{total}\\\text{children}&0.14&0.26&0.40\\\text{adults}&0.21&0.39&0.60\\\text{total}&0.35&0.65&1.00\end{matrix}

a. "Chocolate" and "Adults" (whatever those mean) will be independent as long as

P(\text{chocolate}\cap\text{adults})=P(\text{chocolate})\cdot P(\text{adults})

"Chocolate" has the marginal distribution given by the second column, with a total probability of P(\text{chocolate})=0.35. Similarly, "Adults" has the marginal distribution described by the third row, so that P(\text{adults})=0.60. Then

P(\text{chocolate})\cdot P(\text{adults})=0.35\cdot0.60=0.21

Meanwhile, the joint probability of "Chocolate" and "Adults" is given by the cell in the corresponding row/column, with P(\text{chocolate}\cap\text{adults})=0.21.

The probabilities match, so these events are indeed independent.

Parts (b) and (c) are checked similarly.

b. Yes;


P(\text{children})\cdot P(\text{chocolate})=0.40\cdot0.35=0.14
P(\text{children}\cap\text{chocolate})=0.14

c. Yes;

P(\text{vanilla})\cdot P(\text{children})=0.65\cdot0.40=0.26
P(\text{vanilla}\cap\text{children})=0.26
5 0
3 years ago
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