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3 years ago

15

What o angles around point add up to ?

1 answer:

ziro4ka [17]3 years ago

8 0

Angles around a point sum up to 360°

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A rope weighs 400 grams per meter of length. What is the weight in kilograms of 6.3 m of this rope

torisob [31]

<span>A rope weighs 400 grams per meter of length.
length=6.3m

400*6.3=2520g
2520g=2520/1000kg=2.52kg

The answer is D.</span>

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3 years ago

Give some examples of real life situations in which you could use unit rates to solve an equivalent ratio problem

scoray [572]

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3 years ago

Solve the equation above

Roman55 [17]

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3 years ago

Read 2 more answers

19. On Saturday, a middle school tennis team played in a tournament. The ratio of matches won to the total

mestny [16]

D is correct because if you add 19+7=28 so its d

<h2>thank you</h2>

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4 years ago

According to Brad, consumers claim to prefer the brand-name products better than the generics, but they can't even tell which is

max2010maxim [7]

Answer:

a. There is not enough evidence to support the claim that the proportion that correctly identifies the chip is significantly smaller than 50%.

b. Test statistic z=-1.001

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that the proportion that correctly identifies the chip is significantly smaller than 50%.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.5\\\\H_a:\pi$

The significance level is 0.01.

The sample has a size n=199.

The sample proportion is p=0.462.

$p=X/n=92/199=0.462$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5*0.5}{199}}\\\\\\ \sigma_p=\sqrt{0.001256}=0.035$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.462-0.5+0.5/199}{0.035}=\dfrac{-0.035}{0.035}=-1.001$

This test is a left-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=P(z$

As the P-value (0.16) is greater than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion that correctly identifies the chip is significantly smaller than 50%.

7 0

4 years ago