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JulsSmile [24]
4 years ago
6

5/10 x 9/12 will give brainliest if correct

Mathematics
2 answers:
mote1985 [20]4 years ago
3 0

Answer:

0.375

Step-by-step explanation:

yarga [219]4 years ago
3 0

The answer is \frac{3}{8}

First, you must find like-denominators.

\frac{30}{60} *\frac{45}{60}

Then, multiply the terms across.

\frac{1350}{3600}

Finally, simplify the fraction

\frac{3}{8}

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ladessa [460]

Answer:

-2/1

Step-by-step explanation:

i hope this helps :)

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3 years ago
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HACTEHA [7]

Answer:

the anwer is B ( i mean second option)

And you can try it

you will find ;

y =  \frac{x}{3}  - 1

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Step-by-step explanation:

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3 years ago
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svp [43]

Answer:1

Step-by-step explanation:

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3 years ago
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What is the inverse function for h(x) = <img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B7%7D%20-1" id="TexFormula1" title="\sqrt
Blizzard [7]

Step-by-step explanation:

y =  \sqrt{x + 7}  - 1

Replace x and y.

x =  \sqrt{y + 7}  - 1

Solve for y.

x + 1 =  \sqrt{y + 7}

Square both sides.

(x + 1) {}^{2}  = y + 7

(x + 1) {}^{2}  - 7 = y

This is the inverse function or you can also say

{x}^{2}  + 2x - 6 = y

5 0
3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
3 years ago
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