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Korolek [52]
2 years ago
6

PLZ HELP WILL GIVE BRAINLIST!!!

Mathematics
2 answers:
yulyashka [42]2 years ago
5 0

Answer:

Purple is 9%

Teal is 12%

Yellow is 4%

FromTheMoon [43]2 years ago
3 0

Answer:

purple=9%

teal=12%

yellow=4%

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Maria and Franco are mixing sports drinks for a track meet. Maria uses 23 cup of powdered mix for every 2 gallons of water. Fran
Anastaziya [24]

Answer:

Maria

Step-by-step explanation:

Find the ratio of powdered mix to water for each drink.

For Maria:

(2/3 cup) / (2 gallon)

= 1/3 cup per gallon

For Franco:

(1 1/4 cup) / (5 gallon)

= (5/4 cup) / (5 gallon)

= 1/4 cup per gallon

Since 1/3 is bigger than 1/4, Maria's sports drink is stronger.

4 0
2 years ago
(n/‒7) + 3 = ‒2<br> explain it
iris [78.8K]

Answer:

The answers is n=35

Step-by-step explanation:

Solve for n by simplifying both sides of the equation. Then isolating the variable.

J’espère que ça aide.

5 0
2 years ago
Read 2 more answers
Find the slope of the line through the given points (9,2) and (7,8)
Kipish [7]
The velocity, slope, of a line is always (y2-y1)/(x2-x1), in this case:

slope=(8-2)/(7-9)=6/-2=-3

m=-3
3 0
3 years ago
Jill said "Eight times my age 3 years ago equals 104?How old is Jill now
OleMash [197]

12.625, so about 13.

8 0
2 years ago
Read 2 more answers
1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The avera
ioda

Answer:

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

                \frac{Xbar - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = $1080

                s  = sample standard deviation = $260

                 n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, \mu is given by;

P(-2.032 < t_3_4 < 2.032) = 0.95

P(-2.032 < \frac{Xbar - \mu}{\frac{s}{\sqrt{n} } } < 2.032) = 0.95

P(-2.032 * {\frac{s}{\sqrt{n} } < {Xbar - \mu} < 2.032 * {\frac{s}{\sqrt{n} } ) = 0.95

P(X bar - 2.032 * {\frac{s}{\sqrt{n} } < \mu < X bar + 2.032 * {\frac{s}{\sqrt{n} } ) = 0.95

95% confidence interval for \mu = [ X bar - 2.032 * {\frac{s}{\sqrt{n} } , X bar + 2.032 * {\frac{s}{\sqrt{n} } ]

                                            = [ 1080 - 2.032 * {\frac{260}{\sqrt{35} } , 1080 + 2.032 * {\frac{260}{\sqrt{35} } ]

                                             = [ 990.70 , 1169.30 ]

<em>No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.</em>

Therefore, the store manager believe is not correct.

8 0
3 years ago
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