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3 years ago

10

If there is 1/10 of the recommended daily intake of iron in 3/4 cup of oatmeal, what fraction of the recommended daily intake of

iron is in 1 cup of oatmeal?

2 answers:

rewona [7]3 years ago

8 0

Answer:

Step-by-step explanation:

elena55 [62]3 years ago

3 0

Answer:

2/15

Step-by-step explanation:

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A gallery has 25 paintings in its permanent collection, with display space for 10 at one time. How many different collections ca

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You need to know in how many ways you can choose 10 out of 25 items when order does not matter. This is a combination problem.

$_{n}C_{r} = \dfrac{n!}{(n - r)!r!}$

$_{25}C_{10} = \dfrac{25!}{(25 - 10)!10!}$

$_{25}C_{10} = \dfrac{25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15!}{15! \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$_{25}C_{10} = \dfrac{25 \times 24 \times 23 \times 22 \times 21 
\times 20 \times 19 \times 18 \times 17 \times 16}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 
\times 2 \times 1}$

$_{25}C_{10} = \dfrac{23 \times 11 \times 20 \times 19 \times 2 \times 17}{1}$

$_{25}C_{10} = 23 \times 11 \times 20 \times 19 \times 2 \times 17$

$_{25}C_{10} = 3,268,760$

6 0

3 years ago

Read 2 more answers

A supervisor records the repair cost for 14 randomly selected refrigerators. A sample mean of $79.20 and standard deviation of $

Nitella [24]

Answer:

( $74.623, $83.777)

The 90% confidence interval is = ( $74.623, $83.777)

Critical value at 90% confidence = 1.645

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $79.20

Standard deviation r = $10.41

Number of samples n = 14

Confidence interval = 90%

Using the z table;

The critical value that should be used in constructing the confidence interval.

z(α=0.05) = 1.645

Critical value at 90% confidence z = 1.645

Substituting the values we have;

$79.20+/-1.645($10.42/√14)

$79.20+/-1.645($2.782189528308)

$79.20+/-$4.576701774067

$79.20+/-$4.577

( $74.623, $83.777)

The 90% confidence interval is = ( $74.623, $83.777)

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WILL PICK BRAINLIEST PLZ ANSWER

Mars2501 [29]

Answer:

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