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2 years ago

75 DIVIDE BY 600 EQUALS

Mathematics

1 answer:

ad-work [718]2 years ago

4 0

Answer:

here?

Step-by-step explanation:

75÷600=0.125

if you meant

600÷75=8

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GEOMETRY <br> please Help

geniusboy [140]

Answer:

A right circular cone

Step-by-step explanation:

7 0

2 years ago

The first terms of a geometric sequence are 0.778,-2.33,7,-21, and 63. What is the 6th term?

aliya0001 [1]

Answer:

-189

Step-by-step explanation:

If you continue the pattern by multiplying the term by -3, you will eventually get -189 as the 6th term

3 0

3 years ago

Help please

mezya [45]

Answer:

Step-by-step explanation:

Remark

My guess is that what is confusing you is not what you have to do, but why it is disguised as g(n)

What you are doing in effect is setting up a table. You are also not certain where the table starts. And that is a problem. I will start it at zero, but it might be 1.

zero

n = 0

g(0) = 34 - 5*0

g(0) = 34

One

n = 1

g(1) = 34 - 5*1

g(1) = 34 - 5

g(1) = 29

Two

g(2) = 34 - 5*2

g(2) = 34 - 10

g(2) = 24

Three

g(3) = 34 - 5*3

g(3) = 34 - 15

g(3) = 19

Four

g(4) = 34 - 5*4

g(4) = 34 - 20

g(4) = 19

Answer

0 1 2 3 4

34 29 24 19 14

7 0

2 years ago

You will need the same dataset used for problem 3 in homework 1 (the dataset obtained from the yahoo website with the company yo

stepladder [879]

Answer:

a) The probability of sampling at random a fish that is smaller in size than the value you would obtain by subtracting half the standard deviation from the average is 0.3085.

b) The probability of sampling at random a fish that is greater in size than the value you would obtain by adding half the standard deviation from the average is 0.3085.

c) The probability of sampling at random a fish that has a size between the two values is 0.383.

d) The 25th and 75 percentiles of fish size for the population using the normal distribution table is 5.69 and 5.87 respectively.

e) The probability that the average calculated will be less than the value is 0.3707.

Step-by-step explanation:

For the given data set mean $(\mu) = 5.75913$

Standard deviation $(\sigma) = 0.172229$

Variance $(\sigma2) = 0.0296$

Here we get is

$P(x < \mu - \sigma/2) = p(x < 5.673) \\\\ = 0.3085$

$P(x < \mu + \sigma/2) = p(x < 5.845) \\\\= 0.3085$

$P(\mu - \sigma/2 < x < \mu + \sigma/2) = p(5.673 < x < 5.845) \\\\= 0.383$

25th percentile:-

$= 25*[(n+1)/100]th term \\\\= 5.69$

75the percentile:-

$= 75*[(n+1)/100]th term\\\\ = 5.87$

$p(x < \mu - \sigma/3) = p(x < 5.7017) \\\\= 0.3707$

7 0

2 years ago

Q1.Simplify:

solniwko [45]

$\textbf{(i)}\\\\\{1^3 +2^3 \} \times \left(\dfrac 13 \right)^2\\\\=(1+8) \left(\dfrac 19 \right)\\\\=9\left(\dfrac 19 \right)\\\\=1\\\\$

$\textbf{(ii)}\\\\\{5^{-1}\times 4^{-1}\}^2\\\\=\left(5^{-1} \right)^2 \times \left(4^{-1} \right)^2~~~~~~~~~~~;[(ab)^m = a^mb^m]\\\\=5^{-2}\times 4^{-2}~~~~~~~~~~~~~~~~~~~~;[(a^m)^n = a^{mn}]\\\\=\dfrac 1{5^2} \times \dfrac 1{4^2}~~~~~~~~~~~~~~~~~~~~~~;\left[a^{-m} = \dfrac 1{a^m},~ a\neq 0 \right]\\\\=\dfrac{1}{400}\\\\=0.0025\\\\$

$\textbf{(iii)}\\\\\left\{ \left(\dfrac 13 \right)^{-2} \times \left( \dfrac 12 \right)^{-2}\right\}\div \left(\dfrac 14 \right)^{-3}\\\\\\=\left( \dfrac 13 \times \dfrac 12 \right)^{-2} \div\left(4^{-1}\right)^{-3}\\\\\\=\left(\dfrac 16 \right)^{-2} \div 4^3\\\\\\=\left(6^{-1} \right)^{-2}\div 4^3\\\\\\=6^2 \div 4^3\\\\\\=36\div 64\\\\\\=\dfrac{9}{16}\\\\\\=0.5625$

6 0

2 years ago

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