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3 years ago

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

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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

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

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