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2 years ago

What is an inequality for 20 and 1.29?

Mathematics

1 answer:

Anton [14]2 years ago

4 0

Answer:

Step-by-step explanation:

I think . . .

20 > 1.29

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8% of what number is 2 :)

Alisiya [41]

8% of what number is 2 - 268832

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2 years ago

Read 2 more answers

Simplify: 18q2−45q+25/9q2−25.

Svet_ta [14]

Answer:

$\frac{(6q-5)}{(3q+5)}$

Step-by-step explanation:

Given: $\frac{18q^{2}-45q+25 }{9q^{2}-25 }$

Factorizing the numerator and the denominator, we have:

18 $q^{2}$ - 45q + 25 = 18 $q^{2}$ - 30q - 15q + 25

= (3q - 5)(6q - 5)

and

9 $q^{2}$ - 25 = (3q - 5)(3q + 5)

So that,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(3q-5)(6q-5)}{(3q-5)(3q+5)}$

= $\frac{(6q-5)}{(3q+5)}$

Therefore,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(6q-5)}{(3q+5)}$

5 0

2 years ago

Explain I don’t understand

Serggg [28]

So just divide 100 by 25 which equals 4 then divide 120 by 30 which also equals to 4 and if the rest equal 4 then its proportional (1)
And then same for 2 and 3 just divided the bottom by the top and if they all get the same answer they are proportion and if one is not the same then it’s not

For the grids if the dots are line diagonally straight and it starts at the 0 then it’s proportional if not then it’s not

Hope this helps :)

7 0

3 years ago

The number of hours spent per week on household chores by all adults has a mean of 28 hours and a standard deviation of 7 hours.

Alchen [17]

Answer:

Probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75 is 0.89435.

Step-by-step explanation:

We are given that the number of hours spent per week on household chores by all adults has a mean of 28 hours and a standard deviation of 7 hours.

Also, sample of 49 adults is given.

Let X = number of hours spent per week on household chores

So, assuming data follows normal distribution; X ~ N( $\mu=28,\sigma^{2}=7^{2}$ )

The z score probability distribution for sample mean is given by;

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

where, $\mu$ = mean hours = 28

$\sigma$ = standard deviation = 7 hours

n = sample of adults = 49

Let $\bar X$ = sample mean hours spent per week on household chores

So, probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75 is given by =P( $\bar X$ > 26.75 hours)

P( $\bar X$ > 26.75 hours) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{26.75-28}{\frac{7}{\sqrt{49} } }$ ) = P(Z > -1.25) = P(Z < 1.25)