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Alexus [3.1K]
3 years ago
13

Who now that I do now is cus I am in 5 grade

Mathematics
1 answer:
vovangra [49]3 years ago
5 0

Answer:

The answer is 33/5 I will explain below

Step-by-step explanation:

There are two steps to this first you multiply your whole number which is 6 by the denominator or bottom number in your fraction which is the 5.

Then the second step would be to add your numerator or top number in your fraction to the answer you got from the first step. You would get 33 after step 2

Then you basically keep the the denominator and you end up with 33/5!

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34 petri dishes are prepared with a growth medium. 16 are randomly selected and treated with antibiotic A. The other 18 are trea
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Answer:

(a) According to the central limit theorem, the distributions of the sample means of sufficiently large samples randomly selected from a population with mean, μ and standard deviation, σ with replacement will be normally distributed

Therefore, given that the size of the population from which the samples were selected (34 petri dishes) is comparable the sizes of the samples, (16 and 18), therefore, the samples are approximately normal

Also given that the petri dishes were prepared with growth medium designed to increase the growth of microorganisms, with an expected amount of growth, the samples therefore came from approximately normal distributions

Step-by-step explanation:

8 0
3 years ago
Help me with this please! 10 points
finlep [7]

{  y    ≥   2x   +   1;    y      >       -2x     -   3 } Unbound



Check out the Uploads!!!




Hope that helps!!!!                        : )

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7 0
3 years ago
How do you work out how many 1/3 cup servings are in 6 cups of pecans
allochka39001 [22]

Answer:

18 servings

Step-by-step explanation:

Because 6 divided by 1/3 is the same as 6*3, we can just do it that way so it makes more sense.

6*3= 18

And there you have your answer!

6 0
3 years ago
Read 2 more answers
If 25% of all students in the 7th grade have pets, and there are 40 students with pets, how many total students are there in the
VashaNatasha [74]

Answer:

there would be 160 as if 25% of the student body is 40 then you would multiply it by 4 to get to 100% of the student body thus giving you 160.

Step-by-step explanation:

4 0
3 years ago
g In a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained
Marat540 [252]

Answer:

95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

Step-by-step explanation:

We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.

Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;

                          P.Q. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of residents 65-years or older who had obtained a flu shot = \frac{28}{111} = 0.25

          n = sample of residents 65-years or older = 111

          p = population proportion of residents who were getting the flu shot

<em>Here for constructing 95% confidence interval we have used One-sample z test for proportions.</em>

<u>So, 95% confidence interval for the population proportion, p is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } },\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

   = [ 0.25-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } } , 0.25+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } } ]

   = [0.169 , 0.331]

Therefore, 95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

7 0
3 years ago
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