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3 years ago

Translate eighty-eight less than m is smaller than −176

Mathematics

2 answers:

xeze [42]3 years ago

6 0

Answer:

$88$

Step-by-step explanation:

Katarina [22]3 years ago

3 0

Answer:

88 < m < -176

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3 years ago

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng

alexandr402 [8]

Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.

b. A 95% confidence interval for $\mu$ .

Answer:

$\= x = 13.4$ . $E = 2.3$

$10.7 < \mu < 16.1$

Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The 90% confidence interval is (11.1, 15.7)

Generally the point estimate of $\mu$ is mathematically evaluated as

$\= x = \frac{11.1 + 15.7 }{2}$

=> $\= x = 13.4$

Generally the margin of error is mathematically evaluated as

$E = \frac{15.7 - 11.1}{2 }$

=> $E = 2.3$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the equation for the lower limit of the confidence interval is

$\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1$

=> $13.4 - 0.3877 s = 11.1$

=> $s = 5.932$

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 normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * \frac{5.932}{\sqrt{18} }$

=> $E = 2.7$

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $13.4 - 2.7 < \mu < 13.4 + 2.7$

=> $10.7 < \mu < 16.1$

3 0

3 years ago

Help please asap today is the deadline

Korvikt [17]

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3 years ago

The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 30 mm and standard d

skelet666 [1.2K]

Answer:

$z=-0.842$

And if we solve for a we got

$a=30 -0.842*7.8=23.432$

So the value of height that separates the bottom 20% of data from the top 80% is 23.432.

Step-by-step explanation:

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(30,7.8)$

Where $\mu=30$ and $\sigma=7.8$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.80$ (a)

$P(X$ (b)

As we can see on the figure attached the z value that satisfy the condition with 0.20 of the area on the left and 0.80 of the area on the right it's z=-0.842

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.842$

And if we solve for a we got

$a=30 -0.842*7.8=23.432$

So the value of height that separates the bottom 20% of data from the top 80% is 23.432.

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3 years ago

Put<br> 4y+x=-1 into y-intercept form and show how you did it.

artcher [175]

4y+x= -1
-x -x
—————
4y=-x-1
— ——
4 4
=======
Y= -1/4x - 1/4
Hoped this helps

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4 years ago