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Tamiku [17]
3 years ago
15

A pet store has three aquariums for fish. Aquarium A contains 24 fish, Aquarium B contains 51 fish, and Aquarium C contains 36 f

ish. Which of the following graphs would represent this data most accurately? The y-axis of a bar graph starts at zero fish. One bar is 24 units, another bar is 51 units, and the third bar is 36 units. A pictograph shows three cubes. One cube has a side length of 8, another cube has a side length of 17, and the third cube has a side length of 12. The y-axis of a bar graph starts at 20 fish. One bar is 24 units, another bar is 51 units, and the third bar is 36 units. A pictograph shows three squares. One square has a side length of 2.4, another square has a side length of 5.1, and the third cube has a side length of 3.6. ill give brainliest to whoever actually answers first!!!!!
Mathematics
1 answer:
bazaltina [42]3 years ago
5 0

The graph that can represent the data most accurately is (a) The y-axis of a bar graph starts at zero fish. One bar is 24 units, another bar is 51 units, and the third bar is 36 units

The given parameters are:

  • Aquarium A: 24 fishes
  • Aquarium B: 51 fishes
  • Aquarium C: 36 fishes

The above dataset can be represented on a bar graph, where the lengths of the bars represent the number of fishes in each aquarium

Hence, the graph that can represent the data most accurately is (a)

Read more about graphs at:

brainly.com/question/25677468

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If f(x) = 2x^2 + 1 and g(x) = x^2 - 7, find (f - g)(x).
zavuch27 [327]
2x squared +1 (-)x squared -7 is
a.) x squared +8
6 0
3 years ago
Read 2 more answers
The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.08
kvv77 [185]

Answer:

a) 44.93% probability that there are no surface flaws in an auto's interior

b) 0.03% probability that none of the 10 cars has any surface flaws

c) 0.44% probability that at most 1 car has any surface flaws

Step-by-step explanation:

To solve this question, we need to understand the Poisson and the binomial probability distributions.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Poisson distribution with a mean of 0.08 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

So \mu = 10*0.08 = 0.8

(a) What is the probability that there are no surface flaws in an auto's interior?

Single car, so Poisson distribution. This is P(X = 0).

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493

44.93% probability that there are no surface flaws in an auto's interior

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?

For each car, there is a p = 0.4493 probability of having no surface flaws. 10 cars, so n = 10. This is P(X = 10), binomial, since there are multiple cars and each of them has the same probability of not having a surface defect.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003

0.03% probability that none of the 10 cars has any surface flaws

(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws?

At least 9 cars without surface flaws. So

P(X \geq 9) = P(X = 9) + P(X = 10)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 9) = C_{10,9}.(0.4493)^{9}.(0.5507)^{1} = 0.0041

P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003

P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0041 + 0.0003 = 0.0044

0.44% probability that at most 1 car has any surface flaws

5 0
3 years ago
Ellus
butalik [34]

9514 1404 393

Answer:

  2.25

Step-by-step explanation:

Add the square of half the x-coefficient to complete the square.

  (-3/2)² = 9/4 = 2.25

4 0
3 years ago
Automobile warranty claims for engine mount failure in a troppo malo 2000 se are rare at a certain dealership, occurring at a me
Irina-Kira [14]
The probability that the dealership will wait at least 6 months until the next claim is solved by:We know that the mean number of occurrences per time unit is 0.1, then let’s denote this by λ
So: f (x) = 1 – e ^ -λx for x > 0
P (x > 6) = e^ -x
= e ^-0.1 * 6
= 0.5488
4 0
3 years ago
A random sample of 10 observations was selected from a normal population distribution. The sample mean and sample standard devia
Delvig [45]

Answer:

18.0167≤x≤21.9833

Step-by-step explanation:

Given the following

sample size n = 10

standard deviation s = 3.2

Sample mean = 20

z-score at 95% = 1.96

Confidence Interval = x ± z×s/√n

Confidence Interval = 20 ± 1.96×3.2/√10

Confidence Interval = 20 ± (1.96×3.2/3.16)

Confidence Interval = 20 ± (1.96×1.0119)

Confidence Interval = 20 ±  1.9833

CI = {20-1.9833, 20+1.9833}

CI = {18.0167, 21.9833}

Hence the required confidence interval is 18.0167≤x≤21.9833

8 0
3 years ago
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