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GenaCL600 [577]

2 years ago

9

Please help solve -5<14-4x≤3

1 answer:

Anna35 [415]2 years ago

4 0

Answer:

Interval notation (-5,3]

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Is the relation a function? Why or why not?<br><br> {(–3, –2), (–1, 0), (1, 0), (5, –2)}

tankabanditka [31]

Answer:

yes, the given relation is a function.

Step-by-step explanation:

The given relation is

{(–3, –2), (–1, 0), (1, 0), (5, –2)}

A relation is called function if each element of the domain is paired with exactly one element of the range.

It means for each value of x there exist a unique value of y.

In the given relation for each value of x there exist a unique value of y.

Therefore the required solution is yes and this relation is a function.

5 0

3 years ago

Five less than the product of 3 and a number of 40 . Write in an equation

vagabundo [1.1K]

Answer:

I don't get the "number of 40" part, but here's my best shot!

3(40x)-5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

0<br>3. Find the area of a circle with a diameter of 6 inches

Xelga [282]

Answer:

28.26

Step-by-step explanation:

To find the area of a circle you take pi times the radius squared ( $\pi r^{2}$ )

And to find the radius you divide the diameter by two.

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3 years ago

IMG_9606.JPG Help on all of it plzz

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8 0

3 years ago

A randomized trial tested the effectiveness of diets on adults. Among 36 subjects using Diet 1, the mean weight loss after a yea

seropon [69]

Answer:

The 95% confidence interval is

$0.45 < \mu_1 - \mu_2 < 5.35$

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 36$

The first sample mean is $\= x_1 = 3.5$

The first standard deviation is $\sigma_1 = 5.9 \ pounds$

The second sample size is $n_2 = 36$

The second sample mean is $\= x_2 = 0.6$

The second standard deviation is $\sigma = 4.4$

Generally the degree of freedom is mathematically represented as

$df = \frac{ [ \frac{s_1^2 }{n_1 } + \frac{s_2^2 }{n_2} ]^2 }{ \frac{1}{(n_1 - 1 )} [ \frac{s_1^2}{n_1} ]^2 + \frac{1}{(n_2 - 1 )} [ \frac{s_2^2}{n_2} ]^2 }$

=> $df = \frac{ [ \frac{5.9^2 }{34 } + \frac{4.4^2 }{34} ]^2 }{ \frac{1}{(34 - 1 )} [ \frac{5.9^2}{34} ]^2 + \frac{1}{(34- 1 )} [ \frac{4.4^2}{ 34} ]^2 }$

=> $df =63$

Generally the standard error is mathematically represented as

$SE = \sqrt{ \frac{s_1 ^2 }{n_1} + \frac{s_2^2 }{ n_2 } }$

=> $SE = \sqrt{ \frac{ 5.9 ^2 }{ 36 } + \frac{ 4.4^2 }{36} }$

=> $SE = 1.227$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{\frac{\alpha }{2}, 63 } = 1.998$

Generally the margin of error is mathematically represented as

$E = t_{\frac{\alpha }{2}, 63 } * SE$

=> $E = 1.998 * 1.227$

=> $E = 2.45$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 - \x_2) -E < \mu$

=> $(3.5 - 0.6) - 2.45 < \mu_1 - \mu_2 < ( 3.5 - 0.6) + 2.45$

=> $0.45 < \mu_1 - \mu_2 < 5.35$

8 0

2 years ago