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3 years ago

O is the center. AB||CD. Find y. (a) 24.5° (b) 110° (c) 60° (d) 55°

Mathematics

1 answer:

Leno4ka [110]3 years ago

5 0

Answer:

(d) 55°

Step-by-step explanation:

Because segment AB is parallel to CD, ∠OAB will also have measure y°. (It is an alternate internal angle with ∠COA created by transversal AO.)

We assume that arc AB, marked "70", has a measure of 70°, which means ∠AOB = 70°.

ABO is an isosceles triangle, so its base angles are equal and the sum of its internal angles is 180°.

... y° + y° + 70° = 180°

... 2y = 110

... y = 55

_____

Comment on units

The number 70 is not marked as degrees, so we could take it as either an angle measure or an actual length (in which case the problem cannot be worked). A unitless angle measure is radians, but 70 radians makes no sense in this context. (70 radians is 11 full circles plus about 50.7°. "y" would have to be a rather large negative number for the angle measures to sum to 180° in triangle AOB.)

The value of y we found above is 55, so y° is then 55°. It makes no sense in this context for y to be called 55°, because the angle measure would then be ...

... 55°°

a value that has no meaning as an angle measure.

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A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio

yaroslaw [1]

Answer:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

$\hat p=0.45$ estimated proportion of chips that fail in the first 1000 hours of their use

$\mu_0 =0.48$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion si less then 0.48:

Null hypothesis: $p\geq 0.48$

Alternative hypothesis: $p < 0.48$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v = P(Z$

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4 years ago

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natka813 [3]

Answer: 24

Step-by-step explanation:

you add 12 to 39 which equals 51 then you subtract 51 from 75 which equals 24

hope this helps

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