Juan baked 250 cookies in 2 hours, How many cookies can he make in 1 minute?

baherus [9]

Answer

1• 2/5÷7/8

if there is a division sign in between two fractions what u have to do is multiply and, the fraction at the right need to change

=2/5×8/7

=16/35

=0•457

2• 3/4÷6/5

=3/4×5/6
18/24
=0.75

3• 3/5÷7/8

=3/5×8/7

=24/35

=0•6857

4• 3/4÷5/6

=3/4×6/5

=18/20

=0•9

8 0

3 years ago

The graph below shows the distribution of students grades on their last math test. Based on the Empirical rule, what percent of

timurjin [86]

Based on the emperical rule
the distribution will be as follows

btwn 80 n 85 on the test - 34%
btwn 80 n 95 on the test - 82%
below 85 on the test - 50%
above an 80 on the test - 84%

5 0

4 years ago

Sleep apnea is a disorder in which there are pauses in breathing during sleep. People with this condition must wake up frequentl

Pachacha [2.7K]

Answer:

a) 0.24356 or 24.36%

b) [102.39 , 105.61]

c) The interpretation of this confidence interval is that in samples of 427 people aged 65 years and older, there is a 99% probability that the number of people that suffers sleep apnea is between 103 and 105.

Step-by-step explanation:

a)

This can be considered a binomial distribution (a person either has sleep apnea or not).

Based on the sample we have the probability of suffering the condition is

p = 104/427 = 0.24356

and q (the probability of not suffering the condition) is

q=1-0.24356=0.75644

So the proportion of people aged 65 years and older who have sleep apnea is 0.24356 or 24.36%

b)

To check if we can approximate this binomial distribution with the Normal distribution we must see that

np ≥ 5 and nq ≥ 5

where n is the sample size. Since

427*0.24356 ≥ 5 and 427*0.75644 ≥ 5

we can approximate the binomial with a Normal distribution with mean

np = 427*0.24356 = 104

and standard deviation

$\large s=\sqrt{npq}=\sqrt{427*0.24356*0.75644}=8.867$

The 99% confidence interval (without the continuity correction factor) is given by the interval

$\large [\bar x-z^*\frac{s}{\sqrt n}, \bar x+z^*\frac{s}{\sqrt n}]$

where

$\large \bar x$ is the sample mean

s is the sample standard deviation

n is the sample size

$\large z^*$ is the 0.01 (99%) upper critical value for

the Normal distribution N(0;1).

The value of $\large z^*$ can be found either by using a table or a computer to find it equals to

$\large z^*=2.576$

and our 99% confidence interval (without continuity correction) is

$\large [104-2.576*\frac{8.867}{\sqrt{427}}, 104+2.576*\frac{8.867}{\sqrt {427}}]=[102.8946,105.1054]$

We can now introduce the continuity correction factor. This should be done because we are approximating a discrete distribution (Binomial) with a continuous one (Normal).

This is simply done by widening the interval in 0.5 at each end, so our final 99% confidence interval is

[102.3946 , 105.6054] = [102.39 , 105.61] rounded to 2 decimal places.

c)

The interpretation of this confidence interval is that in samples of 427 people aged 65 years and older, there is a 99% probability that the number of people that suffers sleep apnea is between 103 and 105.

If another study found a 15% of elderly people suffered sleep apnea, that would mean that in a sample of 427 only 64 would have the condition. Since that number is less than 103 by far, that would give us a reason to doubt about the conditions that framed the study (sample size, sampling method, age of people, etc.)

4 0

3 years ago

The total cost, y, for Leah and x friends to go to an amusement park can be modeled by a

Katarina [22]

Answer:

Range of the function is:

{55.25, 100.50, 145.75, 191.00, 236.25, 281.50}

Step-by-step explanation:

Given that:

Seating capacity of the car = 6

Entrance fee per person = $45.25

Parking fee per car = $10