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4 years ago

Find the LCM of 9,25,36 and 73 ?

Mathematics

2 answers:

Artyom0805 [142]4 years ago

6 0

Step-by-step explanation:

9 = 9 , 25 = 5 , 36 = 36

if it's helpful please tell me

Allisa [31]4 years ago

5 0

Answer: The least common multiple is 65700

Step-by-step explanation:

To begin, find the prime factors of the given numbers, (and you find they have very little in common.)

9=3×3 25=5×5 36=2×2×3×3 73= 73×1 (73 is a prime number)

Only the 3's of 36 overlap with the 3's of 9

To find the LCM you must multiply all the prime factors that don't overlap:

3×3×5×5×2×2×73

If the 73 is changed to 72, the LCM is a somewhat lower number: 1800.

To test these LCM's try dividing by any of the factors or numbers given.

If the quotient is not a whole number, the LCM doesn't include that factor.

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3 years ago

88 Is 20 percent of what number?

ValentinkaMS [17]

Answer:

440

Step-by-step explanation:

You can write the problem statement in math terms as ...

88 = 0.20 × (what number)

To solve this, divide by the coefficient of the variable, 0.20.

88/0.20 = (what number)

440 = (what number)

The number you're looking for is 440.

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Of course you know that 20% = 20/100 = 0.20.

(% is a shorthand way to write /100.)

5 0

3 years ago

The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt

AVprozaik [17]

Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average height = 175 cm

$\sigma$ = population standard deviation = 20 cm

n = sample of men = 40

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $175-1.96 \times }{\frac{20}{\sqrt{40} } }$ , $175+1.96 \times }{\frac{20}{\sqrt{40} } }$ ]

= [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.

3 0

3 years ago

Of the radius of a circle is 17 then the diameter of the circle is...

bonufazy [111]

Diameter is 34
Diameter is double the radius so 17 plus 17 is 34 which is the diameter

6 0

3 years ago

A cell phone company sold 4,000 cell phones in the month of November. In December, the company held its annual sale and sold 22%

makvit [3.9K]

First, move the decimal place from the 22% so it would be .22 multiply that by 4,000 that would be 880 add that to 4,000 so 4,880 50% of that would be .50x4880= 2440 subtract that from 4,880 so 4,880-2,440=2,440

In November they sold 4,000
In December they sold 4,880
In January they sold 2,440

add these together and in the 3 months the company sold 11,320 cell phones. :)

3 0

3 years ago

Read 2 more answers

Find the LCM of 9,25,36 and 73 ?​

Find the LCM of 9,25,36 and 73 ?