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lorasvet [3.4K]
3 years ago
13

The pass mark for a test is 86%. Steve scores 52 out of 61 marks. Does he pass the test?

Mathematics
2 answers:
Marina86 [1]3 years ago
5 0

Answer:

nope

Step-by-step explanation:

52÷61=0.85

0.85x100=85%

he needs 86% so he does not pass

laiz [17]3 years ago
4 0

Answer:

He fails

Step-by-step explanation:

Marks = 52/61

Percentage =  52/61  x 100 ~ 85%

Pass amrk =86%

85% < 86%  

So , he fails

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You kept track of the numbr of minutes you spent on homework for one week. 18, 20, 22, 11, 19, 18, 18 What is range of minutes y
frosja888 [35]

Answer:

your range is 11

Step-by-step explanation:

22-11= 11

4 0
3 years ago
Quadrilateral ABCD is inscribed in circle P as shown. Which statement is necessarily true? A. `"m"/_A +"m"/_B = "m"/_C + "m"/_D
pogonyaev

Opposite angles of an inscribed quadrilateral are supplementary, so it is true that ...

... B. m∠A + m∠C = m∠B + m∠D . . . . . = 180°

3 0
3 years ago
Answer!!!!!Please!!!!Will do anything!!!
Marizza181 [45]
I think it is the first one

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3 0
3 years ago
Read 2 more answers
Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in
Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population variance is given as follows:

\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}

It is provided that:

<em>n</em> = 20

<em>s</em> = 3.9

Confidence level = 95%

⇒ <em>α</em> = 0.05

Compute the critical values of Chi-square:

\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}

\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

4 0
3 years ago
Need math help asap please! :/ thanks
Dominik [7]

The rule to remember about generating the perpendicular family to a line is we swap the coefficients on and x and y, remembering to negate one of them. Then the constant is set directly from the intersecting point.


So we have


y = 3x + 2


-3x + 1y = 2


Swapping and negating gets the perpendiculars; the constant is as yet undetermined.


1x + 3y = constant


Since we want to go through (0,2), we could have just written


x + 3y = 0 + 3(2) = 6


3y = -x + 6


y = (-1/3) x + 2


Third choice


5 0
3 years ago
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