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LenaWriter [7]
3 years ago
6

On Monday, Rachel bought 13 yards of fabric in order to make face masks. Later that day, she purchased another 6 3/4 yards. On W

ednesday she made a few face masks and used 5 1/10 yards of fabric. How much fabric does Rachel have left?
Mathematics
1 answer:
Vladimir [108]3 years ago
5 0

Answer:

14.65

Step-by-step explanation:

You add 13 + 6 3/4 then subtract 5 1/10 from that and you have your answer.

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The mean price of a new laptop computer from Store #1 is $799.99, and they only sell two brands of computers. The mean price of
Sati [7]
<span>                                                     value
Weighted mean average = ---------------------------
                                              number of brands


value = mean price of computers from Store #1 * number of brands from store #1 + mean price of computers from store #2 * number of brands from store #2

value =  $799.99 * 2 +  $679.99 * 4 = $4,319.94

number of brands = 2 + 4 = 6

weighted average = $4,319.94 / 6 = $ 719.99

Answer: option b. $719.99
</span>
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2 years ago
Solve for the solution: 2m+1≥7
QveST [7]

Answer:

m≥3

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
I need help with questions two and three I do not understand them at all
rosijanka [135]
1. D bc switch the 1/3 into 3/1, then 4 time 3/1 = 12. Basically 4 times 3 =12
2. D
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2 years ago
Which one is true!!!
irga5000 [103]
B and C I believe .............
7 0
2 years ago
In a sample of 679 new websites registered on the Internet, 42 were anonymous (i.e., they shielded their name and contact inform
Semmy [17]

Answer:

95% Confidence interval:  (0.0429,0.0791)      

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 679

Number of anonymous websites, x = 42

\hat{p} = \dfrac{x}{n} = \dfrac{42}{679} = 0.0618

95% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96

Putting the values, we get:

0.0618\pm 1.96(\sqrt{\dfrac{0.0618(1-0.0618)}{679}}) = 0.0618\pm 0.0181\\\\=(0.0429,0.0791)

is the required confidence interval for proportion of all new websites that were anonymous.

8 0
3 years ago
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