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Morgarella [4.7K]

3 years ago

8

Measurements of a shot recorded on a videotape during a basketball game are shown. the ball passed through the hoop even though

it barely cleared the hands of the player b who attempted to block it. +

1 answer:

Mariulka [41]3 years ago

3 0

I don’t see a question here. Please post the question or more details so we can help you.

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Heyyyyyy besties plz answer this question correctly and u will get brainliest plz thanks

Levart [38]

Answer:

Nicole is more expansive

Step-by-step explanation:

x for Rhonda replaced by 2 = 21. X for Nicole replaced by 2=24

6 0

3 years ago

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If a box of 100 pencils cost $4.75, what should you expect to pay for 225 pencils?

skad [1K]

Answer:

$ 10.69

Step-by-step explanation:

225 is 2.25 times the 100 pencils. ( 225/100) So all you have to do is multiply 2.25 the price ($4.75) and you get 10.6875, money can only go to the tenth place so you round it to $10.69

<em>Hope this helped and pls mark as brainliest!</em>

<em />

<em>~Luna</em>

6 0

3 years ago

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On a coordinate grid, the coordinates of vertices P and Q for polygon RQRS are P(1, 2) and Q(-1, 2)

Nata [24]

We need to find the length of side PQ of the polygon.
Distance formula:

Using distance formula we get

The length of side PQ of the polygon is 5 units.
I think this is what you were asking..?

8 0

3 years ago

These are the weekly wages paid to staff in a hotel.

Marianna [84]

If I’m not mistaken 174 is your median

5 0

3 years ago

A university researcher wants to estimate the mean number of novels that seniors read during their time in college. An exit surv

lys-0071 [83]

Answer:

Population mean = 7 ± 2.306 × $\frac{2.29}{\sqrt{9} }$

Step-by-step explanation:

Given - A university researcher wants to estimate the mean number

of novels that seniors read during their time in college. An exit

survey was conducted with a random sample of 9 seniors. The

sample mean was 7 novels with standard deviation 2.29 novels.

To find - Assuming that all conditions for conducting inference have

been met, which of the following is a 94.645% confidence

interval for the population mean number of novels read by

all seniors?

Proof -

Given that,

Mean ,x⁻ = 7

Standard deviation, s = 2.29

Size, n = 9

Now,

Degrees of freedom = df

= n - 1

= 9 - 1

= 8

⇒Degrees of freedom = 8

Now,

At 94.645% confidence level

α = 1 - 94.645%

=1 - 0.94645

=0.05355 ≈ 0.05

⇒α = 0.5

Now,

$\frac{\alpha}{2} = \frac{0.05}{2}$

= 0.025

Then,

$t_{\frac{\alpha}{2}, df }$ = 2.306

∴ we get

Population mean = x⁻ ± $t_{\frac{\alpha}{2}, df }$ × $\frac{s}{\sqrt{n} }$

= 7 ± 2.306 × $\frac{2.29}{\sqrt{9} }$

⇒Population mean = 7 ± 2.306 × $\frac{2.29}{\sqrt{9} }$

3 0

3 years ago

Read 2 more answers