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3 years ago

How to simplify 24yz^5/32y^2z

Mathematics

1 answer:

Maurinko [17]3 years ago

8 0

24y*z^5/32y^2*z

24/32 = 3/4
y/y^2 = 1/y
z^5/z = z^4/1

Soo now that they are all simplified the answer is 3z^4/4y

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Mr Dominguez spends three weeks training for a marathon. Every day he runs 1/2 mile more than the day before. If the first day h

barxatty [35]

Answer: 16 miles

Explanation:
I believe it would be 16 miles, if he starts off at 6 miles on day one and goes up 1/2 mile every day, he would be up a mile every other day, basically on day 3 he would be at 7 and day 5 at 8 miles, if the pattern continues he would be at 16 miles by day 21.

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3 years ago

Identify the values of a, b, and c that would be used in the quadratic formula to solve the equation 3x2− 5x + 5 = 0. A. a = 3,

stepan [7]

<span>3x2− 5x + 5 = 0.
a=3 b=-5 c=5

A. a = 3, b = 5, c = 5
B. a = 3, b = −5, c = 5
C. a = 5, b = −5, c = 0
D. a = −3, b = 5, c = −5

answer is B

</span>

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4 years ago

It is 76°f at the 6000-foot level of a mountain, and 49°f at the 12,000-foot level of the mountain. write a linear equation to f

valina [46]

Answer:

$t(x) = 103 -0.0045x$

Step-by-step explanation:

We are given the following in the question:

Let the equation:

$t(x) = a + bx$

be the linear equation that represent temperature at an elevation x.

Temperature at 6000-foot level = 76 f

$76 = a + 6000b$

Temperature at 12000-foot level = 49 f

$49 = a + 12000b$

Solving the two equation, we get,

$76-49 = -6000b\\27=-6000b\\\\b = \dfrac{-27}{6000}\\\\b = -0.0045\\76 = a + (-0.0045)6000\\76 = a -27\\a = 76 + 27\\a =103$

Thus, we can write the linear equation as:

$t(x) = 103 -0.0045x$

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3 years ago

container a and b hold 11,875 liters. container b holds 2,391 more liters than container a. how much is in container a?

SIZIF [17.4K]

9484 liters in container A

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3 years ago

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per nig

Kitty [74]

Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let $\mu$ denotes the average hours of sleep per night.

As per given , we have

$H_0:\mu=7\\H_a:\mu$

, since $H_a$ is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : $\overline{x}=7.24$

Sample standard deviation : s= 1.93

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

i.e. $t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58$

For significance level $\alpha=0.05$ and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= $t_{\alpha, df}=t_{0.05,21}=- 1.7207$

Decision : Since the test statistic value (0.58) > critical value 1.7207, it means we are failed to reject the null hypothesis .

[Note : When $|t_{cal}|>|t_{cri}|$ , then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

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3 years ago