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3 years ago

A cheetah ran 300 feet in 2.92 seconds. What was the cheetah’s average speed in miles per hour? Show your work.

Mathematics

1 answer:

alexdok [17]3 years ago

8 0

Answer:

about 70.0 mph

Step-by-step explanation:

speed = distance/time

(300 ft)/(2.92 s)·(1 mi)/(5280 ft)·(3600 s)/(1 h) = (300·3600/(2.92·5280)) mi/h

≈ 70.0498 mi/h

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What is the value of x? Show all of your work.Round your answer to the nearest tenth.

Mashutka [201]

Answer:

$x=20.6\ in$

Step-by-step explanation:

we know that

Applying the Pythagoras Theorem

$41.2^{2} =35.7^{2} +x^{2}$

solver for x

$x^{2}=41.2^{2}-35.7^{2}$

$x^{2}=422.95$

$x=20.6\ in$

8 0

3 years ago

there are 49 dogs signed up for the dog show. there are 36 more small dogs than large dogs. how many small dogs are signed up fo

pogonyaev

Okay, so total dogs are 49.
So we know if x=large dogs and y=small dogs, then x+y=49.
Next we are told that there are 36 MORE small dogs than large dogs. We can take that more meaning addition, and x being our value for large dogs. So y=x+36.
After that we now know what value we can plug in for y. So x+x+36=49.
We can then simplify it to 2x+36=49.
Subtracting the 36 from both sides, leaves you with 2x=13.
Followed by dividing both sides by 2, gives you x=6.5. Or 6.5 large dogs.
Now we can plug this into our formula for the small dogs (y=x+36) to give us y=6.5+36, which simplifies to y=42.5. Or 42.5 small dogs. Which is our answer.
We can double check it by adding the small and large dogs together, 6.5+42.5, which gives us 49, our total entries.

4 0

3 years ago

Read 2 more answers

Assume that body masses of Goldfinch birds follow a normal distribution with standard deviation equal to 0.04 oz. An ornithologi

il63 [147K]

Answer:

The formula to generate 70% confidence interval is: $[\overline{x} - 0.013, \overline{x} + 0.013]$ , in which $\overline{x}$ is the sample mean.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.7}{2} = 0.15$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.15 = 0.85$ , so Z = 1.037.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Assume that body masses of Goldfinch birds follow a normal distribution with standard deviation equal to 0.04 oz.

This means that $\sigma = 0.04$ .

Sample of 10 birds:

This means that $n = 10$ .

The margin of error is of:

$M = z\frac{\sigma}{\sqrt{n}}$

$M = 1.037\frac{0.04}{\sqrt{10}}$

$M = 0.013$

The lower end of the interval is the sample mean of $\overline{x}$ subtracted by M.

The upper end of the interval is the sample mean of $\overline{x}$ added to M.

Then, the formula to generate 70% confidence interval is: $[\overline{x} - 0.013, \overline{x} + 0.013]$ , in which $\overline{x}$ is the sample mean.

8 0

3 years ago

If f(x) = 5x^2– 3 and f(x + a) = 5x^2+ 30x + 42,<br> what is the value of a ?

Andrei [34K]

Answer:

The zeros are x=0,3,-2

There is a multiplicity of 1 for all of them.

Step-by-step explanation:

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3 years ago

There are five students of different ages. The median age is 13 years. What are two possibilities for the ages of the three olde

Xelga [282]

13 and 14 becuase 13 being the third largest and 14 being the second largest

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3 years ago