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krek1111 [17]
3 years ago
11

2) Mackenzie just started training for a marathon. According to her plan,

Mathematics
2 answers:
IrinaVladis [17]3 years ago
8 0
25/19 = 1.31589474
.0131589474%
raketka [301]3 years ago
6 0

Answer:

32%

Step-by-step explanation:

25 - 19 = 6

6 divided by 19 = 0.315

round to the nearest percent, 32%

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Please help!!! 100 points if correct!!! please actually answer
rosijanka [135]

Answer:

see below

Step-by-step explanation:

What is the theoretical probability that the family has two dogs or two cats? (1/2)

The choices are dd, dc, cd, cc

There are 4 choices = dd or cc/ total = 2/4 = 1/2

Let the heads of the coin be dogs and the tails of a coin be cats

Flip two coins and coin A is the first pet and coin B is the second pet

heads , heads = 10

heads, tails  14

tails heads  =13

tails tails =  13

          total 50

Experimental probability  2 dogs or 2 cats = ( hh, tt) /total = ( 10+13) /50 =                    23/50

If we had 3 pets  

what is the theoretical probability that they have three dogs or three cats?

ddd, ddc, dcd, dcc, ccc, ccd ,cdc, cdd

There are 8 options

 ddd or ccc/ total = 2/8 = 1/4

Let the heads of the coin be dogs and the tails of a coin be cats

Flip three coins and coin A is the first pet and coin B is the second pet Coin C be the third pet

4 0
3 years ago
Eight avocados cost 4 how much does 11 avocados cost
Elden [556K]
It costs 5,5! Bc 1 costs 0,5
5 0
3 years ago
Read 2 more answers
When a fair coin is flipped, what is the chance of getting about 50% heads- specifically between 40% and 60% heads for n=10 flip
aleksklad [387]
At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
3 0
3 years ago
All Questions are fron percentage and its applications
Likurg_2 [28]

Answer:

1.20%

Step-by-step explanation:

1.50-30=20

2.15000

7 0
3 years ago
The bad debt ratio for a financial institution is defined to be the dollar values of loans defaulted divided by the total dollar
Nimfa-mama [501]

Answer:

(a) NULL HYPOTHESIS, H_0 : \mu \leq  3.5%

    ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5%

(b) We conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Step-by-step explanation:

We are given that a random sample of seven Ohio banks is selected.The bad debt ratios for these banks are 7, 4, 6, 7, 5, 4, and 9%.The mean bad debt ratio for all federally insured banks is 3.5%.

We have to test the claim of Federal banking officials that the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

(a) Let, NULL HYPOTHESIS, H_0 : \mu \leq  3.5% {means that the the mean bad debt ratio for Ohio banks is less than or equal to the mean for all federally insured banks}

ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5% {means that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks}

The test statistics that will be used here is One-sample t-test;

                T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where,  \bar X = sample mean debt ratio of Ohio banks = 6%

             s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} } = 1.83%

             n = sample of banks = 7

So, test statistics = \frac{6-3.5}{\frac{1.83}{\sqrt{7} } }  ~ t_6

                             = 3.614

(b) Now, at 1% significance level t table gives critical value of 3.143. Since our test statistics is more than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Hence, Federal banking officials claim was correct.

7 0
3 years ago
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