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xxTIMURxx [149]
3 years ago
9

The prime factorizations of 88 and 92 are shown.

Mathematics
2 answers:
bekas [8.4K]3 years ago
7 0

Answer:

2 x 2 x 2 x 11 x 23

Step-by-step explanation:

 have a good day be safe

OLga [1]3 years ago
4 0

\boxed{LCM=2024}

<h2>Explanation:</h2>

The Least Common Multiple of two integers, say, a and b is the smallest positive integer that is divisible by both a and b. It is usually denoted by LCM(a, b). From the statement we know that the prime factorization of 88 and 92 are:

88 = 2\times 2 \times 2 \times 11 \\ \\ 99 = 2 \times 2 \times 23

So we can find the LCM by taking the common terms with the greatest exponent and the non-common terms, so:

LCM=2\times 2\times 2 \times 11 \times 23 \\ \\ \boxed{LCM=2024}

<h2>Learn more:</h2>

LCM: brainly.com/question/10415148

#LearnWithBrainly

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The current in a river is 1.0 meters/second. Britney swims 300 meters against the current. If she normally swims with a speed of
AVprozaik [17]

Answer:

D) 300 seconds

Step-by-step explanation:

if the river is traveling 1 m/s and Britney is traveling 2 m/s, then half of her speed would be canceled out, resulting in her only traveling 1 m/s taking her 300 seconds or 5 minutes to complete the trip.

5 0
2 years ago
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an account starts with $500 and is compounded quarterly at 5% interest how much is the account worth after 10 years
Tomtit [17]

Answer:

$1500

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 20%/100 = 0.2 per year,

then, solving our equation

I = 500 × 0.2 × 10 = 1000

I = $ 1,000.00

The simple interest accumulated

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at a rate of 20% per year

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5 0
3 years ago
Quarters are currently minted with weights normally distributed and having a standard deviation of 0.065. New equipment is being
Ilya [14]

Answer:

Yes, the new equipment appear to be effective in reducing the variation of​ weights.

Step-by-step explanation:

We are given that Quarters are currently minted with weights normally distributed and having a standard deviation of 0.065.

A simple random sample of 25 quarters is obtained from those manufactured with the new​ equipment, and this sample has a standard deviation of 0.047.

Let \sigma = <u><em>standard deviation of weights of new equipment.</em></u>

SO, Null Hypothesis, H_0 : \sigma \geq 0.065      {means that the new equipment have weights with a standard deviation more than or equal to 0.065}

Alternate Hypothesis, H_A : \sigma < 0.065      {means that the new equipment have weights with a standard deviation less than 0.065}

The test statistics that would be used here <u>One-sample chi-square</u> test statistics;

                           T.S. =  \frac{(n-1)s^{2} }{\sigma^{2} }  ~ \chi^{2}__n_-_1

where, s = sample standard deviation = 0.047

           n = sample of quarters = 25

So, <u><em>the test statistics</em></u>  =  \frac{(25-1)\times 0.047^{2} }{0.065^{2} }  ~  \chi^{2}__2_4   

                                     =  12.55

The value of chi-square test statistics is 12.55.

Now, at 0.05 significance level the chi-square table gives critical value of 13.85 at 24 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of chi-square as 12.55 < 13.85, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that the new equipment have weights with a standard deviation less than 0.065.

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3 years ago
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Y_Kistochka [10]
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Indicate the equation of the given line in standard form.
Minchanka [31]

wait do you want it in standard form

5 0
3 years ago
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