All categories

3 years ago

What percent of the first 10 natural numbers are prime numbers? Answer in a percent please. :)

1 answer:

6 0

So the first 10 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Since a prime number is only divisible by 1 and itself and greater than 1, our answers are 2, 3, 5, and 7. That's 4 numbers, or 4/10=0.4 of the 10. Since percentage is based out of 1, we can multiply 0.4 by 100 to get 40%. I hope that Helped :)

You might be interested in

In an instant lottery, your chances of winning are 0.1. if you play the lottery five times and outcomes are independent, the pro

lesya692 [45]

The probability that you lose all five times will be 0.59.

<h3>How to calculate the probability?</h3>

From the information given, the chances of winning are 0.1. Therefore, the chance of losing will be:

= 1 - 0.1 = 0.9

Therefore, the probability that you lose all five times will be:

= 0.9 × 0.9 × 0.9 × 0.9 × 0.9

= 0.59

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

6 0

2 years ago

A real estate agent sells a house for $92,000. She receives a 6% commission on the sale of the house.

denis23 [38]

Okay so just do this equation below.

92,000 x 0.06

And you have your answer!

Answer is 5,520.

Leave a 5 star rating, a thanks, and a most brainy answer!

5 0

3 years ago

Read 2 more answers

Which picture shows the correct graph of |x|=3?

suter [353]

Answer:

[see picture]

Step-by-step explanation:

There were no given choices, but I did graph the equation using Desmos.

Choose the answer that is closest to this:

8 0

3 years ago

Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$

5 0

1 year ago

George desposited 2000 in his bank, which offers 5% compoud interest annually. What would be the principla amount available in g

Tamiku [17]

George would have 2200

4 0

3 years ago