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3 years ago

If 40% of a given number is 8, then what is 15% of the given number?

1 answer:

5 0

40% of a given number is 8...
0.40x = 8
x = 8 / 0.40
x = 20....so 20 is the given number

what is 15% of 20...
0.15(20) = 3 <==== ur answer

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Estimate.then find your product 92x68

Nonamiya [84]

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Answer (product): 6256

7 0

3 years ago

Read 2 more answers

A boy who is flying a kite lets out 300 feet of string which makes an angle of 60o with the ground. Assuming that the string is

Aleks [24]

<h3>Answer :</h3>

See the attachment for better understanding.

Given : Length of string (AC) = 300ft

To Find : Height of the kite from ground (AB)

$:\implies\sf\:sin60^{\circ}=\dfrac{AB}{AC}$

$:\implies\sf\:\dfrac{\sqrt{3}}{2}=\dfrac{H}{300}$

$:\implies\sf\:H=\dfrac{300\sqrt{3}}{2}$

$:\implies\sf\:H=150\sqrt{3}$

$:\implies\:\boxed{\bf{\red{H\approx 260\:feet}}}$

Hope it helps !

6 0

3 years ago

Drag steps in the given order to evaluate this expression. (-5)(2) – 2(-3) + 3

sesenic [268]

Answer:

-1

Step-by-step explanation:

using PEDMAS(parentheses, exponents, division, multiply, addition, subtraction) to solve the problem

First, multiply (-5)(2) and 2(-3)

(-5)(2) – 2(-3) + 3

=-10-(-6)+3

=-10+6+3

add -10+6

-4+3

Add -4+3

-4+3

=-1

Therefore, (-5)(2) – 2(-3) + 3 is equal to -1

4 0

3 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

3 years ago

A group of 15 explorers to the south pole had just enough food to last them for 40 days how long would the food last if all of a

Trava [24]

Answer:

3 days

Step-by-step explanation:

15-3=12

40/12= 3.3

round to 3

5 0

3 years ago