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Vadim26 [7]
3 years ago
12

The science class is taking a field trip. There are a 20 students and a unknown number of adults. The bus holds 35 people at mos

t. What is the greatest number of adults who can chaperone?
Mathematics
2 answers:
AveGali [126]3 years ago
5 0
It would be 15 because the 35 people at most minus the 20 students equals 15 so that would be the greatest number of adults allowed to be there.
Monica [59]3 years ago
5 0
15 because 35 minus 20 equals 15.
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A car travels 45km in an hour. In each of the next two hours, it travels 78km. What is the average speed of the car.
romanna [79]

Answer:

The average speed of car is 67 kilometers per hour.

Step-by-step explanation:

Given that:

Distance travelled in one hour = 45 km

Distance travelled in next 2 hours = 78*2 = 156 km

Total distance = 45+156 = 201

Total time = 1+2 =  3 hours

Average speed of the car = \frac{Total\ distance}{Total\ time}

Average speed of the car = \frac{201}{3}

Average speed of the car = 67 km/hr

Hence,

The average speed of car is 67 kilometers per hour.

7 0
3 years ago
Profit is the difference between revenue and cost. The revenue, in dollars, of a company that manufactures televisions, can be m
Irina-Kira [14]

Answer:

50,700

Step-by-step explanation:

7 0
2 years ago
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A racquetball club membership costs $22.50 per month. What would a year’s membership cost?
dsp73

Answer:

270 dollars for a year, so 22.50 times 12

8 0
3 years ago
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3n+75=50+2n please help righting more
alexandr1967 [171]

Answer: n = -25

Step-by-step explanation:

you would need to get N by itself so you would subtract 2n from both sides and that would bring you to n+75=50 and then you would need to subtract 75 from both sides and that leads you to n=-25

5 0
3 years ago
A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y de
strojnjashka [21]

Answer:

a) P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

a= μ-3.16*σ , b= μ+3.16*σ

b) P(Y≥ μ+3*σ ) ≥ 0.90

b= μ+3*σ

Step-by-step explanation:

from Chebyshev's inequality for Y

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

where

Y =  the number of fish that need be caught to obtain at least one of each type

μ = expected value of Y

σ = standard deviation of Y

P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean

k= parameter

thus for

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

P{a≤Y≤b} ≥ 0.90 →  1-1/k² = 0.90 → k = 3.16

then

P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

using one-sided Chebyshev inequality (Cantelli's inequality)

P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)

P{Y≥b} ≥ 0.90  →  1- σ²/(σ²+λ²)=  1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ

then for

P(Y≥ μ+3*σ ) ≥ 0.90

5 0
3 years ago
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