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IRINA_888 [86]
1 year ago
5

What -7 x4=36? amogus amogus amogus amogus

Mathematics
1 answer:
77julia77 [94]1 year ago
3 0

Based on the given problem of -7x⁴ = 36, the solution to the equation is x = -1.5.

<h3>What is the solution to this problem?</h3>

The equation to find x is given as:

-7x⁴ = 36

Solving it gives:

-7x⁴ = 36

x⁴ = 36/-7

x⁴ = -36/7

x = ⁴√(-36/7)

x = -1.5

In conclusion, x is -1.5.

Find out more questions on solving for x at brainly.com/question/2910769

#SPJ1

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Step-by-step explanation:

\frac{1}{3x}  -  \frac{1}{2}  = 18 \frac{1}{2}  \\  \\  \frac{1}{3x}  -  \frac{1}{2}  =  \frac{37}{2}   \\  \\  \frac{1}{3x}  =  \frac{37}{2}  -  \frac{1}{2}  \\  \\  \frac{1}{3x}  =  \frac{37 - 1}{2}  \\  \\  \frac{1}{3x}  =  \frac{36}{2}  \\  \\  \frac{1}{3x}  = 18 \\  \\ 1 = 54x

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<h2>Answer with explanation:</h2>

Let \mu be the population mean.

By considering the given information , we have

Null hypothesis : H_0: \mu=73.4

Alternative hypothesis :  H_a: \mu>73.4

Since alternative hypothesis is right-tailed , so the test is a right-tailed test.

Given : Sample size : n=16 , which is a small sample , so we use t-test.

Sample mean: \overline{x}=78.3  ;

Standard deviation: s=8.4

Test statistic for population mean:

t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}

i.e. t=\dfrac{78.3-73.4}{\dfrac{8.4}{\sqrt{16}}}\approx2.333

Using the standard normal distribution table of t , we have

Critical value for \alpha=0.01 : t_{(n-1,\alpha)}=t_{(15,0.01)}=2.602

Since , the absolute value of t (2.333) is smaller than the critical value of t (2.602) , it means we do not have sufficient evidence to reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the claim that answering questions while studying produce significantly higher exam scores.

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Step-by-step explanation:

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