All categories

3 years ago

In a survey, 250 adults and children were asked whether they know how to

Mathematics

2 answers:

Andreas93 [3]3 years ago

8 0

Hello, I'm Eric. I'll be trying my best to assist you on your question today.

0.12 is 12 percent.

We also have the 0.06 percent which would just be 6 percent

We need to add up those percentages to get 18 percent, for the people surveyed that cannot swim would be 18 percent.

Our final answer is D.

Not the right answer or confused? Reply to this question for help.

Enjoy your day - Eric

Artist 52 [7]3 years ago

7 0

Answer:

Step-by-step explanation:

B because .06 + .12 = .18 which is 18% out of 250 people

You might be interested in

Q.1 Simplify :- (i) (- 625 ) + 925 + 100 + (-200)

Dmitrij [34]

825 - 625i

Hope it’s right
Best luck with your studying

3 0

2 years ago

40 and 400 The value of 4 in ___ is______ times the value of 4 in _____

Nat2105 [25]

The value of 4 in the 400 is ten times the value of 4 in the 40's place

7 0

3 years ago

Read 2 more answers

A shoe store recorded the price (in dollars) of 175 pairs of women's shoes and 175 pairs of men's shoes. The data collected was

arsen [322]

Answer:

Most of the men's shoes in the data set cost more than women's shoes.

Step-by-step explanation:

From what I could tell it can't be The price of men's shoes has a higher interquartile range than that of women's shoes. becuse that's false what looks to fit is Most of the men's shoes in the data set cost more than women's shoes. since on the graph more of the mens vaules are bigger

5 0

3 years ago

At a baseball game, a vender sold a combined total of 240 sodas and hot dogs. The number of sodas sold was two times the number

olga55 [171]

Answer:

Sodas: 160

Hot Dogs: 80

Step-by-step explanation:

Sodas: s

Hot Dogs: d

s+d=240

s=2d

3d=240

d=80

s=160

7 0

3 years ago

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

3 0

3 years ago