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GaryK [48]
1 year ago
9

A pet store has 15 birds and 75 fish.

Mathematics
2 answers:
Ludmilka [50]1 year ago
6 0

Answer:

1/5

Step-by-step explanation:

Alex Ar [27]1 year ago
6 0

Answer: b

Step-by-step explanation: 1/5

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Is the equation x9 – 5x3 + 6 = 0 quadratic in form? Explain why or why not.
podryga [215]

Answer:

NOO

Step-by-step explanation:

In order for an equation to be quadratic, the highest degree that is needed is two. The highest exponent provided is 3 therefore no it is not a quadratic function

6 0
2 years ago
Read 2 more answers
Select the correct answer. Which transformation will always map a parallelogram onto itself? A. a 90° rotation about its center
prisoha [69]

Answer:

C. a 180° rotation about its center

Step-by-step explanation:

The question is on rotation of an object about the origin through 180°

When a point B, (h,k) is rotated about the origin 0 through 180° both clockwise and anticlockwise directions , the new position is B'(-h,-k)

Example, if we have point K (5,8) rotated through 180°, clockwise or anti-clockwise about the origin the new position will be; K'(-5,-8)

7 0
1 year ago
Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
2 years ago
An increase in walking has been shown to contribute to a healthier life-style. A sedentary American takes an average of 5000 ste
Zielflug [23.3K]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 5000

For the alternative hypothesis,

H1: µ > 5000

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 5000

x = 5430

σ = 600

n = 40

z = (5430 - 5000)/(600/√40) = 4.53

Looking at the normal distribution table, the probability corresponding to the z score is < 0.0001

Since alpha, 0.05 > than the p value, then we would reject the null hypothesis. Therefore, at a 5% level of significance, it can be concluded that they walked more than the mean number of 5000 steps per day.

6 0
1 year ago
How should the headings of a résumé be formatted so that they are clear and easy to find? a. bold and italicized b. capitalized
agasfer [191]

Answer:

The Heading should be Bold and Capital

Step-by-step explanation:

LIKE THIS :)

4 1
2 years ago
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