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Alenkinab [10]
3 years ago
14

For the past 40 days, Naomi has been recording the number of customers at her restaurant between 10:00 a.m. and 11:00 a.m. Durin

g that hour, there have been fewer than 20 customers on 25 out of the 40 days.
What is the experimental probability there will be **20 or more customers** on the forty-first day? Include it as fraction, decimal and percent.
Mathematics
1 answer:
olganol [36]3 years ago
3 0

Answer:

the experimental probability for 20 or more customersis 37.5%

Step-by-step explanation:

Given that

The recording of the customers would be in between 10:00 AM to 11:00 PM

In this, they have been fewer than 20 customers on 20 out of the 40 days

We need to find out the experimental probability for 20 or more customers

So,

= (40 - 25) ÷ (40)

= 15 ÷ 40

= 37.5%

Hence, the experimental probability for 20 or more customers is 37.5%

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Write a linear function in slope intercept form only that corresponds with this table​
Bezzdna [24]
Y = 3x + 1

Because,
slope intercept form is y=mx+b
Where y is the y values, m is slope, x is x values, and b is y-intercept

First, to find the slope use rise/run, so rise (from 10 to 13) is 3 and run is (from 3 to 4) is 1, 3/1=3

Next to find y-intercept plug in the slope and some x and y values,
10=3(3)+b
10=9+b
1=b
So then the equation is y=3x+1 you can check by plugging stuff in and also I graphed it so I’ll add a picture

This is my first time answering questions so if I got something wrong ask or something hope this helps :)

3 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
Verizon [17]
A) 0.9803; 0.4803
B) 32

We calculate the z-score for this problem by using the formula:

z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}

Using our formula, we have:

z=\frac{3.00-2.65}{\frac{0.85}{\sqrt{25}}}=2.06

Using a z-table (http://www.z-table.com) we see that the area to the left of, less than, this score is 0.9803.

To find the probability it is between the mean and this, we subtract the probability associated with the mean (0.5) from this:
0.9803 - 0.5 = 0.4803.

To find B, we first find the z-score for this.  Using a z-table (http://www.z-table.com) we see that the closest z-score would be 2.33.  We then set up our equation as

2.33=\frac{3.00-2.65}{\frac{0.85}{\sqrt{n}}}=\frac{0.35}{\frac{0.85}{\sqrt{n}}}
\\
\\2.33=0.35\div \frac{0.85}{\sqrt{n}}=0.35\times \frac{\sqrt{n}}{0.85}

Multiplying both sides by 0.85 we have
2.33(0.85) = 0.35√n
1.9805 = 0.35√n

Divide both sides by 0.35:
1.9805/0.35 = √n

Square both sides:
(1.9805/0.35)² = n
32 ≈ n
8 0
3 years ago
Which expression is equivalent to 60 – 3y - 9?
spayn [35]

Answer:

your answer is A

Step-by-step explanation:

:)

3 0
3 years ago
3 (x-4) =<br> 5 (3x+4) =
Veronika [31]

Step-by-step explanation:

3 times x and - 4 and 5 times 3x and 5 easy man

8 0
3 years ago
Read 2 more answers
Eight swimmers participate in a race in how many ways can the swimmers finish in first second and third place an it be answered
Vesnalui [34]

Answer:

336

Step-by-step explanation:

To define whether we use permutations or combinations we must define whether or not the order in which we arrange the results matters. If the order matters we use permutations and if the order doesn't matter we use combinations.

In this case, since we are talking about the first places in a competition, order definitely does matter, so we use permutations. Also in permutations it must be indicated if repetition is allowed, in this case not because the same person cannot be in more than one place.

We use the following formula:

P=\frac{n!}{(n-r)!}

where in this case n is the numer of swimmers and r is the number of places we are considering (1st 2nd and 3rd), which is 3 places.

n = 8

and

r = 3

thus the number of permutations is given bt:

P=\frac{8!}{(8-3)!}\\ \\P=\frac{8!}{5!} \\\\P=\frac{8*7*6*5*4*3*2*1}{5*4*3*2*1}\\ \\P=8*7*6\\P=336

the answer is that there are 336 ways in which the swimmers can finish in first second and third place

3 0
3 years ago
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