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3 years ago

PLEASE HELP!!! Shade 30% of the whole grid. How much blocks do i shade?

Mathematics

1 answer:

Semenov [28]3 years ago

6 0

Answer:

Shade 6

Step-by-step explanation:

There are 20 squares in the grid. Multiply 0.30 (percentages are written as decimals) by 20. There's your answer.

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Foil Multiplication (2n+4)•(4n+7)

kiruha [24]

(2n+4)x(4n+7)

= 8n^2+14x+16n+28
=8n^2+30n+28

that's it :) !!

5 0

3 years ago

Read 2 more answers

{c(1) = 3/16<br> {c(n)=c(n−1)⋅4<br> <br><br> what is the 3rd term in this sequence?

mote1985 [20]

Answer:

$\large\boxed{c(3)=3}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}c(1)=\dfrac{3}{16}\\\\c(n)=c(n-1)\cdot 4\end{array}\right$

Put n = 2 and next n = 3 to the recursive formula:

$c(2)=c(2-1)\cdot4=c(1)\cdot4\to c(2)=\dfrac{3}{16}\cdot4=\dfrac{3}{4}\\\\c(3)=c(3-1)\cdot4=c(2)\cdot4\to c(3)=\dfrac{3}{4}\cdot4=3$

4 0

3 years ago

What is the difference between lcm and gcf and all of that help im doing it on my home work pls help i hate this stuff best answ

raketka [301]

greatest common factor=gcf

least common multule=lcm

with 12 and 16

factors of 12=2 times 2 times 3

factors of 16=2 times 2 times 2 times 2

greates common factor is 2 times 2=4

least common multiple means all factors combined minus repeats or 2 times 2 times 2 times 2 time 3=48

4 0

3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago

Expand the following: 3(16x-24)

Fynjy0 [20]

Answer:

48x-72

Step-by-step explanation:

u multiply 3 by 16x then multiply 3 by 24 and bring down your minus sign

6 0

3 years ago