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CaHeK987 [17]
2 years ago
15

How can i convert 4/9 into a decimal

Mathematics
2 answers:
rosijanka [135]2 years ago
5 0

Answer:

4/9 in a decimal is 0.44444444

tamaranim1 [39]2 years ago
3 0
4/9 in a decimal is .44444444 repeating
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Write expression that is equivalent to 1/4 a -3
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4-2=2 that is the answer to
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2 years ago
F the radius of a flying disc is 7.6 centimeters, what is the approximate area of the disc?
hoa [83]

Answer:

3.14x7.6=23.864 because raidus is the disctance around  one circle

Step-by-step explanation:

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3 years ago
2. Given f(x) = 4x -2 and g(x) = x2 + 1, evaluate:
Novosadov [1.4K]

Answer:

a).

f(5) = 18

b).

g(-3) = 10

c).

g(x-2) = x²- 3

d).

f(g(x)) = 4x²- 1

3 0
2 years ago
A recent national survey indicated that 73 percent of respondents try to include locally grown foods in their diets. A 95 percen
lilavasa [31]

Answer:

Correct option is:

"Less than 75% of all people in the country try to include locally grown foods in their diets."

Step-by-step explanation:

A hypothesis for the proportion of people who include locally grown foods in their diets can be defined as:

<em>H₀</em>: The proportion of people who include locally grown foods in their diets is <em>p</em>.

<em>Hₐ</em>: The proportion of people who include locally grown foods in their diets is difference from <em>p</em>.

The decision rule can be based on the 95% confidence interval.

If the confidence interval consists the hypothetical value of the true proportion then the null hypothesis will be accepted or else the null hypothesis will be rejected.

The 95% confidence interval for the proportion of all people in the country who try to include locally grown foods in their diets is given as (0.70, 0.76).

This interval implies that there is 0.95 probability that the true proportion of people who include locally grown foods in their diets is between 70% and 76%.

Assuming that the claim made was supported by the 95% confidence interval.

The claim should be:

Less than 75% of all people in the country try to include locally grown foods in their diets.

Because the 95% confidence interval defines the range of the true proportion as 70% to 76%. Most of the values are below 75%.

Thus, the correct option is:

"Less than 75% of all people in the country try to include locally grown foods in their diets."

3 0
3 years ago
Read 2 more answers
Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

Z = \frac{X - \mu}{\sigma}

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

5 0
2 years ago
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