This is from my Sociology class and there's not that option soooo

Free brainless answer pleaseeee and morw

sergejj [24]

1. alternate interior angles, x= -10
2. corresponding angles, x= 30

7 0

3 years ago

An assembly line has two inspectors. The probability that the first inspector will miss a defective item is 0.06. If the defecti

Novay_Z [31]

Answer:

probability that both passes a defective item is 0.8742‬

<h3>Step-by-step explanation:</h3>

probability that the first inspector misses is Pr( 1st misses)= 0.06

therefore the probability he does not miss is

Pr(1st passes)= 1 - Pr( 1st misses) = 1 - 0.06 = 0.94

probability that the second misses is Pr( 2nd misses) = 0.07

therefore probability that 2nd does not miss is

Pr( 2nd passes) = 1- Pr( 2nd misses) = 0.93

probability that both passes a defective item is Pr(1st passes)*Pr( 2nd passes)

= 0.93*0.94 = 0.8742‬

7 0

3 years ago

Find the surface area of the pyramid shown to the nearest whole number. 6ft 5ft 5ft

Tom [10]

Answer:75,

Step-by-step explanation:

1/2 bh= A for one T 5*5=25 25*6= 150

150/2=75

OR

5*5/2=12.5

12.5*6=75

4 0

2 years ago

jennifer scored 75 on her quiz last week. this week she scored 90. what was the percent of increase on the quiz score?

Arlecino [84]

Answer:

The answer is a 20% increase

Hope this helps!

Mark me brainliest if I'm right :)

4 0

2 years ago

A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco

sammy [17]

Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

Let p = proportion of all cars in San Francisco who are blue

SO, Null Hypothesis, $H_0$ : p = 25% {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, $H_A$ : p $\neq$ 25% {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of 600 cars in San Francisco who are blue = $\frac{141}{600}$ = 0.235

n = sample of cars = 600

So, test statistics = $\frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }$

= -0.866

Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

4 0

3 years ago