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Gwar [14]
3 years ago
8

1.) Write a quadratic rule that represents the table below in standard form.

Mathematics
1 answer:
kvv77 [185]3 years ago
3 0

Step-by-step explanation:

1st quadrant is(+,+)

2nd quadrant is(+,-)

3rd quadrant is (-,-)

4th quadrant is (-,+)

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Fin the domain of the function f(x)=x^2+7
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Answer:

all real numbers

Step-by-step explanation:

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2 years ago
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What is the value of the expression 10 + (fraction 1 over 2)4 ⋅ 48?
djverab [1.8K]
10 + 1/2*4*48 => 10 + 2*48 => 10 + 96 = 106.
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3 years ago
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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So \mu = 373, \sigma = 67

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So n = 5, s = \frac{67}{\sqrt{5}} = 29.96

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 373}{29.96}

Z = 1.23

Z = 1.23 has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So \mu = 526, \sigma = 107

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So n = 5, s = \frac{107}{\sqrt{5}} = 47.86

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 526}{47.86}

Z = -2.42

Z = -2.42 has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0
3 years ago
Drinking 8 fluid ounces of milk provides 270.0 milligrams of calcium. How many fluid ounces of milk provide 72.5 milligrams of c
Margarita [4]
Set up a proportion:

\frac{8}{270} = \frac{x}{72.5}

The numerator represents fluid ounces of milk, and the denominator represents milligrams of calcium.

Simplify the first fraction to make multiplication easier:

\frac{8}{270} \div  \frac{2}{2} =  \frac{4}{135}
\frac{4}{135} = \frac{x}{72.5}

Cross multiply the fractions:

72.5 * 4 = 290
135x = 290

Divide both sides by 135 to get x by itself:

290 \div 135 = 2.1
x = 2.1


2.1 fluid ounces of milk provide 72.5 milligrams of calcium.
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If x equals 6, what is the order of the angles from smallest to largest degree?
aleksandrvk [35]

Answer:

A

Step-by-step explanation:

just find the value , and you will get tje answer

8 0
3 years ago
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