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2 years ago

What are all the factors of 54?

Mathematics

2 answers:

nekit [7.7K]2 years ago

8 0

Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54.

Step2247 [10]2 years ago

5 0

Answer:

1, 2, 3, 6, 9, 18, 27, 54 (answer choice D. for edge students)

Step-by-step explanation:

Factorization deals with listing out the highest multiples of the given number, it's really a matter of knowing your times tables, but informational search engines/websites works just fine as a substitution.

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the lombaro family went out to dinner and the cost of their meal was 62% if they left a 20% tip how much did they leave

swat32

Answer:

they left a $1.24 tip

Step-by-step explanation:

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3 years ago

Surface area is always measured in square units. True False

OLEGan [10]

<h3>Answer: True</h3>

Often we replace "units" with something like "feet" or "meters"

examples: square feet, square meters, square yards, etc

The "square" refers to the idea that any surface area can be unwrapped and lay flat, then you can break the area of that flat figure into smaller squares. A good example of this is a dice has 6 faces and each face is a square itself, so the total surface area is the sum of all the 6 square areas.

6 0

2 years ago

Read 2 more answers

Rewa Delta Union Rugby CEO has become concerned about the slow pace of the rugby games played in the current union rugby, fearin

ozzi

Answer:

We conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

Step-by-step explanation:

We are given that Before the meeting, the mean duration of the 15-sided rugby game time was 3 hours, 5 minutes, that is, 185 minutes.

A random sample of 36 of the 15-sided rugby games after the meeting showed a mean of 179 minutes with a standard deviation of 12 minutes.

Let $\mu$ = mean duration of 15-sided union rugby games after the meeting.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 185 minutes {means that the mean duration of 15-sided union rugby games has increased or remained same after the meeting}

Alternate Hypothesis, $H_A$ : $\mu$ < 185 minutes {means that the mean duration of 15-sided union rugby games has decreased after the meeting}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean duration of 15-sided union rugby games = 179 min

s = sample standard deviation = 12 minutes

n = sample of 15-sided rugby games = 36

So, the test statistics = $\frac{179-185}{\frac{12}{\sqrt{36} } }$ ~ $t_3_5$

= -3

The value of t test statistics is -3.

Now, at 1% significance level the t table gives critical value of -2.437 at 35 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of t as -3 < -2.437, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

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3 years ago

Your gross income is $56k and the standard deduction is $12,200, how much do you owe in taxes?

Neko [114]

8 percent is the answer

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2 years ago

A movie theater is holding a contest for people who are in line before 10 p.m. The people who are within three places of the 99

pav-90 [236]

Answer:

|x-99|<3

Step-by-step explanation:

|x-99|<3

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3 years ago