Pls, help me asap-10 points :)

out of 32 students in a class, 5 said they ride a bike to school. based on these results, how many of the 800 students in the sc

Brut [27]

We can write a proportion for this question
5 out of 32 students and we have to find how many (x) out of 800
5/32 = x /800
cross multiply
5*800 = 32x
4000 = 32x
divide both sides by 32
125 = x

About 125 students of the 800 ride their bikes to school.

Hope this helps :)

8 0

3 years ago

What has to happen for scales to be balanced?

Anon25 [30]

Answer:

When the pans contain exactly the same mass the beam is in balance. ... You can place an object in one pan and standard weights in the other to find what the object weighs. Here balance scales are used to show that the box "x" has a mass of 4

5 0

3 years ago

Read 2 more answers

Help look at the pic i sent in the attachments( it says heloo don't judge ok) i will mark you bainliest and i will give you 12 p

Kisachek [45]

Answer:D

Step-by-step explanation:

7 0

3 years ago

Lesson 6.07

Oduvanchick [21]

Answer:

A) Interval estimate for those that want to get out earlier = (35%) ± (4%) = (31%, 39%)

Interval estimate for those that want to get out later = (39%) ± (4%) = (35%, 43%)

B) The group that wants to get out of school earlier can win after all the votes are counted if their true population proportion takes on a value that is higher than the closest true population proportion (for the group that wants to get out of school later)

That is, in the (31%, 39%) and (35%, 43%) obtained in (a), a range of (35.1%, 39%) and (35%, 38.9%) show how possible that the group that wants to get out of school earlier can win after all the votes are counted.

Step-by-step explanation:

The Interval estimate for the a proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Interval estimate = (Sample proportion) ± (Margin of error)

Margin of Error is the width of the confidence interval about the proportion.

a) Find the interval estimates

i) for those that want to get out earlier and

ii) those that want to get out later.

i) Sample proportion of those that want to get out earlier = 35%

Margin of Error = 4%

Interval estimate for those that want to get out earlier = (35%) ± (4%) = (31%, 39%)

ii) Sample proportion of those that want to get out later = 39%

Margin of Error = 4%

Interval estimate for those that want to get out later = (39%) ± (4%) = (35%, 43%)

b) Explain how it would be possible for the group that wants to get out of school earlier to win after all the votes are counted.

Since the interval estimates represent the range of values that the true population proportion can take on for each group that prefer a particular option, the group that wants to get out of school earlier van have their proportion take on values between 31% and 39%. If their true population takes on a value that is highest (which is very possible from the interval estimate), and the group with the highest proportion in the sample, (the group that wants to get out of school later, whose true population proportion can take between 35% and 43%) has a true population proportion that is less than that of the group that wants to get out of school earlier, then, the group that wants to get out of school earlier can win after all the votes are counted.

Hope this Helps!!!

7 0

3 years ago

Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin w

Goshia [24]

Answer:

Step-by-step explanation:

Given that;

the following procedure for accomplishing our task are:

1. Flip the coin.

2. Flip the coin again.

From here will know that the coin is first flipped twice

3. If both flips land on heads or both land on tails, it implies that we return to step 1 to start again. this makes the flip to be insignificant since both flips land on heads or both land on tails

But if the outcomes of the two flip are different i.e they did not land on both heads or both did not land on tails , then we will consider such an outcome.

Let the probability of head = p

so P(head) = p

the probability of tail be = (1 - p)

This kind of probability follows a conditional distribution and the probability of getting heads is :

$P( \{Tails, Heads\})|\{Tails, Heads,( Heads ,Tails)\})$

$= \dfrac{P( \{Tails, Heads\}) \cap \{Tails, Heads,( Heads ,Tails)\})}{ {P( \{Tails, Heads,( Heads ,Tails)\}}}$

$= \dfrac{P( \{Tails, Heads\}) }{ {P( \{Tails, Heads,( Heads ,Tails)\}}}$

$= \dfrac{P( \{Tails, Heads\}) } { {P( Tails, Heads) +P( Heads ,Tails)}}$

$=\dfrac{(1-p)*p}{(1-p)*p+p*(1-p)}$

$=\dfrac{(1-p)*p}{2(1-p)*p}$

$=\dfrac{1}{2}$

Thus; the probability of getting heads is $\dfrac{1}{2}$ which typically implies that the coin is fair

(b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

For a fair coin (0<p<1) , it's certain that both heads and tails at the end of the flip.

The procedure that is talked about in (b) illustrates that the procedure gives head if and only if the first flip comes out tail with probability 1 - p.

Likewise , the procedure gives tail if and and only if the first flip comes out head with probability of p.

In essence, NO, procedure (b) does not give a fair coin flip outcome.

5 0

3 years ago