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3 years ago

Please help ASAP please need to get this done

Mathematics

1 answer:

Vlad1618 [11]3 years ago

5 0

I'm not sure about first two pictures

the third one is:

1.75

the fourth picture
11 ft^2

5.

960cm ^3

hope this helps

good luck

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Y=1/2x Y= 4x Y= 1/4 x Y= 2 x

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It's a line in form y = mx.

Two points:

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The formula of a slope:

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Substitute:

$m=\dfrac{1-0}{4-0}=\dfrac{1}{4}$

Answer: $y=\dfrac{1}{4}x$

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3 years ago

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3 years ago

Micah is preparing for a piano recital. He practices for 50 minutes on Monday and 90 minutes on Wensday

SCORPION-xisa [38]

Answer:

She increased 80% the practise time

Step-by-step explanation:

The question is incomplete, the following is missing:

by how much did she increase the time she practised each day?

From Monday to Wednesday she increased 90 - 50 = 40 minutes the practise time.

To compute the increment as a percentage use: increment/reference *100

In this case, that is: 40/50*100 = 80%, where 50 minutes is taken as a reference.

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3 years ago

Read 2 more answers

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%