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2 years ago

An object was moving at a rate of 18.2 feet per second for 38.5 seconds. How far did the object travel?

Mathematics

1 answer:

lidiya [134]2 years ago

4 0

Answer:

700.7

Step-by-step explanation:

You're looking for distance. So all you have to do is 18.2 x 28.5 and you'll get your answer.

Hope this helps. :)

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Which system of equations can be used to find the roots of the equation 4x^5-12x^4+6x=5x^3-2x?

Sveta_85 [38]

$^{(1)}\ \ 4x^5-12x^4+6x=5x^3-2x\ \ ^{(2)}\\\\ \left\{\begin{array}{ccc}y=4x^5-12x^4+6x\ \ \ ^{(1)}\\y=5x^3-2x\ \ \ ^{(2)}\end{array}\right$

Answer: system nr 4.

7 0

2 years ago

Jody, a statistics major, grows tomatoes in her spare time. She measures the diameters of each tomato. Assume the Normal model

lina2011 [118]

Answer:

Zscore = 0.5

Step-by-step explanation:

If we assume a normal distribution, we mean that the diameters of each tomato follow a normal distribution. This is N~ (0,1).

By that, we mean that the mean (μ) = 0 and variance ( $\sigma^{2}$ ) = 1. Thus, since we are told that one tomato was in the 50th percentile. This implies the median. And is 0.5. And if the distribution is normal, the mean and median and mode should be equal.

Thus:

==> Z score = $\frac{x-\mu}{\sigma}$ = $\frac{0.5 - \mu}{\sigma} = \frac{0.5-0}{1} = 0.5$

8 0

2 years ago

Jinghua hiked 4 1/2 miles through the woods in 2 1/4 hours. She hiked the return trip at the same average rate but by a differen

Anna007 [38]

Answer:

5 miles.

Step-by-step explanation:

Consider the question: Jinghua hiked 4 1/2 miles through the woods in 2 1/4 hours. She hiked the return trip at the same average rate but by a different route taking 2 1/2 hours. How many miles did Jinghua hike on the return trip ?

First of all, we will find Jinghua's speed using given information as:

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Convert mixed fractions into improper fractions:

$4\frac{1}{2}\Rightarrow \frac{9}{2}$

$2\frac{1}{4}\Rightarrow \frac{9}{4}$

$\text{Jinghua's speed}=\frac{\frac{9}{2}\text{ Miles}}{\frac{9}{4}\text{ Hours}}$

Using property $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}$ :

$\text{Jinghua's speed}=\frac{9*4\text{ Miles}}{9*2\text{ Hours}}$

$\text{Jinghua's speed}=\frac{2\text{ Miles}}{\text{ Hour}}$

We know that distance is equal to the product of speed and time.

$\text{Distance}=\text{Speed}\times\text{Time}$

Since we have been given that Jingua hiked the return trip at the same average rate, so distance covered by her on return trip would be speed (2 miles her hour) times given time (2 1/2 hours).

$\text{Distance covered by Jingua on return trip}=\frac{2\text{ Miles}}{\text{ Hour}}\times 2\frac{1}{2}\text{ Hours}$