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3 years ago

Someone please help me with this

Mathematics

1 answer:

amid [387]3 years ago

7 0

Simliar-AA is your answer

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Help!!!!!!!!!!!!!!!!!!!

Afina-wow [57]

Answer:

for question 1 it is the first one. and for question 2 it is b

Step-by-step explanation:

i did this already

8 0

3 years ago

4.07, 4.070, 4.70, 4.0700 what value is not equivalent.

larisa [96]

4.70, all the others can be simplified to 4.07 by taking away the zeroes after the 7 because they are not significant, The one before the 7 stays because it is significant as it is holding the tenths place value.

5 0

3 years ago

Read 2 more answers

I can not figure out how to do this question and I need help please

spayn [35]

$\bf \textit{quadratic formula}\\\\

\begin{array}{lcclll}
1x^2&-6x&+12&=0\\
\uparrow &\uparrow &\uparrow \\
a&b&c
\end{array}
\qquad \qquad 
x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}$

6 0

3 years ago

Word form for 49.564

d1i1m1o1n [39]

Answer:

49 decimal five hunderd sixty four

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Suppose that at a state college, a random sample of 41 students is drawn, and each of the 41 students in the sample is asked to

ra1l [238]

Answer:

The confidence interval for 90% confidence would be narrower than the 95% confidence

Step-by-step explanation:

From the question we are told that

The sample size is n = 41

For a 95% confidence the level of significance is $\alpha = [100 - 95]\% = 0.05$ and

the critical value of $\frac{\alpha }{2}$ is $Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }= 1.96$

For a 90% confidence the level of significance is $\alpha = [100 - 90]\% = 0.10$ and

the critical value of $\frac{\alpha }{2}$ is $Z_{\frac{\alpha }{2} } =Z_{\frac{0.10 }{2} }= 1.645$

So we see with decreasing confidence level the critical value decrease

Now the margin of error is mathematically represented as

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

given that other values are constant and only $Z_{\frac{\alpha }{2} }$ is varying we have that

$E\ \ \alpha \ \ Z_{\frac{\alpha }{2} }$

Hence for reducing confidence level the margin of error will be reducing

The confidence interval is mathematically represented as

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

Now looking at the above formula and information that we have deduced so far we can infer that as the confidence level reduces , the critical value reduces, the margin of error reduces and the confidence interval becomes narrower

3 0

3 years ago