What's 21% of 1,300 A 2,601 B 1,027 C 62 D 273

1 answer:

8 0

$\displaystyle 21\% \cdot 1300 \\ \\ \\ \frac{21}{100}\cdot130 = \\\\ \\ \frac{21}{1}\cdot 13= \\ \\\\ 21\cdot13= \\ \\ \\ 273 \\ \\ \\ \text{The answer is D.273}$

You might be interested in

If f(x) =15x 2 + 10x, find f(-1) Anyone know the answer?

Rom4ik [11]

I'm assuming the given function is f(x) = 15x^2 + 10x.

If so, then,

f(x) = 15x^2 + 10x
f(x) = 15*(x)^2 + 10*(x)
f(-1) = 15*(-1)^2 + 10*(-1) ... note how x is replaced with -1
f(-1) = 15*(1) + 10*(-1)
f(-1) = 15 - 10
f(-1) = 5

Therefore the answer is 5

7 0

3 years ago

the average number of chocolate chips in a particular brand of chocolate chip chewy cookie is 19. the standard deviation is 2.6.

zzz [600]

The probability illustrated shows that the

percent of cookies would have between 13.8 and 24.2 chips is 95.44%.

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

It is important to note that the requirement to solve the probability between z-values is to know that the probability between the z-values is the difference between the probability of the greatest z-value and the lowest z-value.

This will be illustrated thus:

Z value for 13.8, z =(13.8-19)/2.6 =-2

Z value for 24.2, z =(24.2-19)/2.6 = 2

P( 13.8<x<24.2) =P(-2<z<2)

=P( z <2 –P(z < -2)

=0.9772 -0.0228

=0.9544

= 95.44%

The percent is 95.44%.

Learn more about probability on:

brainly.com/question/24756209

#SPJ1