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3 years ago

Dennis traveled at an average speed of 60 mph on his vacation trip. How far did he travel in the first 4.5 hours

Mathematics

1 answer:

k0ka [10]3 years ago

6 0

Answer: 280 Miles.

Step-by-step explanation: 60 x 4 = 240 + 40 = 280 Miles.

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Mookie Betts of the Boston Red Sox had the highest batting average for the 2018 Major League Baseball season. His average was 0.

Annette [7]

Answer:

Binomial probability, with $n = 5, p = 0.352$

Step-by-step explanation:

For each time Mookie Betts went to bat, there were only two possible outcomes. Either he got a base-hit, or he did not. The probability of getting a hit on each at-bat is independent of any other at-bat. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

His average was 0.352.

This means that $p = 0.352$

Assume he has five times at bat tonight in the Red Sox-Yankees game.

This means that $n = 5$

a. This is an example of what type of probability

Binomial probability, with $n = 5, p = 0.352$

8 0

3 years ago

Round -13.457 to nearest integer

fomenos

Answer:

Answer:

-13

Step-by-step explanation

Rounded to the nearest 0.01 or

the Hundredths Place.

5 0

2 years ago

There are 6 time zones in the United States. The eastern part of the U.S., including New York City, is in the Eastern Time Zone.

VashaNatasha [74]

It’s morning in California

6 0

3 years ago

PLEASE HELP WILL GIVE BRAINLIEST!<br> y= 3x — 5<br><br> y=6x — 8<br> solve with elimination pls

Citrus2011 [14]

Answer:

Point form: (1,-2)

Equation form: x=1 y= -2

Step-by-step explanation:

Hope it helps

4 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago