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tatyana61 [14]
3 years ago
12

PLEASE HELP ME IM BEGGING

Mathematics
1 answer:
anygoal [31]3 years ago
8 0
Stuck on the same thing, tried reviewing and reviewing but now I’ll just try finding the answer key
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A quality control technician works in a factory that produces computer monitors. Each day, she randomly selects monitors and tes
jeka94

Answer:

There is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 300

p = 5% = 0.05

Alpha, α = 0.05

Number of dead pixels , x = 24

First, we design the null and the alternate hypothesis  

H_{0}: p = 0.05\\H_A: p > 0.05

This is a one-tailed(right) test.  

Formula:

\hat{p} = \dfrac{x}{n} = \dfrac{24}{300} = 0.08

z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

Putting the values, we get,

z = \displaystyle\frac{0.08-0.05}{\sqrt{\frac{0.05(1-0.05)}{300}}} = 2.384

Now, we calculate the p-value from excel.

P-value = 0.00856

Since the p-value is smaller than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

5 0
2 years ago
Identify the phase shift of each function. Describe each phase shift ( use a phrase like 3 units to the left):
Temka [501]
ANSWER

Phase shift: 2 units right.

EXPLANATION

The given sine function is

y = - \sin(x - 2)

In mathematics, a horizontal shift may also be referred to as the phase shift.

This function is shifted to the right 2 units.
8 0
3 years ago
I need help plz !!!!
svetlana [45]
X = 66 or 180 - 114, which = 66
3 0
3 years ago
SOMEONE PLEASE HELP!!
german

-12=\dfrac{-12}{1}=\dfrac{12}{-1}=\dfrac{-24}{2}=\dfrac{24}{-2}=\dfrac{-36}{3}=\dfrac{36}{-3}=\dfrac{-48}{4}=\dfrac{48}{-4}=\dots

7 0
3 years ago
How do you simplify this
katen-ka-za [31]
It's 2/5 in decimal form it is 0.4.
5 0
3 years ago
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