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bearhunter [10]
3 years ago
13

Can someone help me with this

Mathematics
1 answer:
dedylja [7]3 years ago
4 0
The is 20 characters long
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Janelys buys milk and apples at the store.
Natali5045456 [20]

Answer:

Each bag costs $4.75

Step-by-step explanation:

41.22 = 3.22 + 8x

38 = 8x

4.75 = x

7 0
2 years ago
75 is the median of the data set 68, 67, 75, 73, 72.
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Truee is right you got it dude
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2 years ago
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You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win $1; if
dsp73

Answer:

Step-by-step explanation:

Given that you and a friend play a game where you each toss a balanced coin.

If the upper faces on the coins are both tails, you win $1;

if the faces are both heads, you win $2;

if the coins do not match (one shows a head, the other a tail), you lose $1 (win (−$1)).

Let Y be the amount won

Then Y can take values as 1,2 and -1

P(Y=1) =P(TT) = 0.25\\P(Y=2) = P(HH) = 0.25\\P(Y=-1) = P(HT)+P(TH) = 0.5

The above is the probability distribution for Y.

6 0
3 years ago
Miko is packing for a trip. The total weight of her luggage cannot exceed 200 pounds. She has 3 suitcases that weigh 57 pounds e
Fittoniya [83]
200+57+12=962. The finale answer that should be shown is 962.
5 0
2 years ago
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A randomized controlled study was designed to test whether regular drinking of cranberry juice can prevent the recurrence of uri
Kamila [148]

Answer:

1. The chi-squared statistic = 10.36

The degrees of freedom = 17

The p-value for the test = 0.89

2. The range of the p-value from the Chi squared table = 0.75 < p-value < 0.90

Step-by-step explanation:

1. The Chi squared test is given as follows;

\chi ^{2} = \sum \dfrac{\left (Observed - Expected  \right )^{2}}{Expected  }

Therefore,

                              UTI   No UTI    %     Total

Cranberry juice       8           42      84     50

Lactobacillus          19          30       61     49

Control                    18          30      60    50

The chi-squared statistic is given as follows;

\chi ^{2} = \dfrac{\left (8- 18\right )^{2}}{18} +  \dfrac{\left (42 - 30\right )^{2}}{30} = 10.36

The chi-squared statistic = 10.36

The degrees of freedom, df = 18 - 1 = 17 since the all of the expected count have a minimum value of 18

With the aid of the calculator we find the p value as p as follows;

p = 0.9 - \dfrac{10.36 - 10.085}{12.972 - 10.085} \times (0.9 - 0.75)

The p-value for the test = 0.89  

2. The range of the p-value from the Chi squared table is given as follows;

0.75 < p-value < 0.90.

5 0
3 years ago
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