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3 years ago

PLEASE HELP TEST DUE SOON WILL GIVE BRAINLIEST AND GOOD REVIEW!!!

Mathematics

1 answer:

swat323 years ago

6 0

Answer:

a, d, e, g

I don't know if it's right :(

this was last year for me :(

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4 years ago

Read 2 more answers

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally dist

Fofino [41]

Answer:

a) $8469-1.645\frac{100}{\sqrt{64}}=8448.4$

$8469+1.645\frac{100}{\sqrt{64}}=8489.6$

So on this case the 90% confidence interval would be given by (8448.4;8489.6)

b) $8469-2.054\frac{100}{\sqrt{64}}=8443.33$

$8469+2.054\frac{100}{\sqrt{64}}=8494.68$

So on this case the 96% confidence interval would be given by (8443.33;8494.68)

Step-by-step explanation:

Assuming this complete problem: On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with s = 100. The composition of the bar has been slightly modified, but the modification is not believed to have affected either the normality or the value of s.

(a) Assuming this to be the case, if a sample of 64 modified bars resulted in a sample average yield point of 8469 lb, compute a 90% CI for the true average yield point of the modified bar. (Round your answers to one decimal place.)

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=8469$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=100$ represent the population standard deviation

n=64 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.645$

Now we have everything in order to replace into formula (1):

$8469-1.645\frac{100}{\sqrt{64}}=8448.4$

$8469+1.645\frac{100}{\sqrt{64}}=8489.6$

So on this case the 90% confidence interval would be given by (8448.4;8489.6)

(b) How would you modify the interval in part (a) to obtain a confidence level of 96%? (Round your answer to two decimal places.)

Since the Confidence is 0.96 or 96%, the value of $\alpha=0.04$ and $\alpha/2 =0.02$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.02,0,1)".And we see that $z_{\alpha/2}=2.054$

Now we have everything in order to replace into formula (1):

$8469-2.054\frac{100}{\sqrt{64}}=8443.33$

$8469+2.054\frac{100}{\sqrt{64}}=8494.68$

So on this case the 96% confidence interval would be given by (8443.33;8494.68)

7 0

3 years ago

What is the equation of the line that passes through the point (-2,14) and is perpendicular to the line with the following equat

tino4ka555 [31]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{5}}x-1\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-2}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{-2}\implies \cfrac{5}{2}}}$

so we're really looking for the equation of a line whose slope is 5/2 and it passes through (-2 , 14)

$(\stackrel{x_1}{-2}~,~\stackrel{y_1}{14})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{5}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{14}=\stackrel{m}{ \cfrac{5}{2}}(x-\stackrel{x_1}{(-2)}) \implies y -14= \cfrac{5}{2} (x +2) \\\\\\ y-14=\cfrac{5}{2}x+5\implies {\Large \begin{array}{llll} y=\cfrac{5}{2}x+19 \end{array}}$

3 0

1 year ago

Question is in the image

slamgirl [31]

Answer:

best i could find hope it helps

Step-by-step explanation:

10.000 1 151,200 1 531, 411 1 1,000,000

6 0

3 years ago