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givi [52]
3 years ago
6

Given that Distance = Speed × time, calcuate the distance traveled when a car goes 55 mi/h for 8 hours.​

Mathematics
2 answers:
Luba_88 [7]3 years ago
6 0
55x8= 440, hope this helps
anastassius [24]3 years ago
4 0

Answer:

440

Step-by-step explanation:

55*8=440

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Find the zeros of the function f(x)=-0.3x^2+0.6x+3.8 to the nearest hundredth
Paul [167]

Ce clasă ești sa văd daca te pot ajuta

6 0
3 years ago
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 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 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

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

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

Z = \frac{410 - 375}{27.76}

Z = 1.26

Z = 1.26 has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

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

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

Z = \frac{410 - 523}{43.27}

Z = -2.61

Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0
3 years ago
By first calculating the size of angle LMN,
Naily [24]

Step-by-step explanation:

Hope it helps.... that is how I did it

4 0
3 years ago
Solve 3x2 + 4x = −5.
Alla [95]
Move the -5 over by adding 5 to both sides
3x^2+4x+5=0
must use quadratic formula

for an equation in the form
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}

a=3
b=4
c=5

x=\frac{-4+/- \sqrt{4^2-4(3)(5)} }{2(3)}
x=\frac{-4+/- \sqrt{16-60} }{6}
x=\frac{-4+/- \sqrt{-44} }{6}
x=\frac{-4+/- 2\sqrt{-11} }{6}
remember that√-1=i
x=\frac{-4+/- 2i\sqrt{-11} }{6}
x=\frac{-2+/- i\sqrt{-11} }{3}

x=\frac{-2+ i\sqrt{-11} }{3} or x=\frac{-2- i\sqrt{-11} }{3}


4 0
3 years ago
What is the answer to this equation
joja [24]

Answer:

S = 9.9

Step-by-step explanation:

First, multiply by -3 to remove fractions:

so:

S - 38.4 = -28.5

Take all values to one side by adding 38.4 to both sides

S = 9.9

5 0
2 years ago
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