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3 years ago

Solve the formula t=v+k/g for g

1 answer:

8 0

$t=v+\frac{k}{g}\\\\v+\frac{k}{g}=t\ \ \ \ |subtract\ v\ from\ both\ sides\\\\\frac{k}{g}=t-v\iff\frac{g}{k}=\frac{1}{t-v}\ \ \ \ |multiply\ both\ sides\ by\ k\neq0\\\\\boxed{g=\frac{k}{t-v}}$

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(2 x 106) x (0.00009) = ?

RSB [31]

I think you meant to write

(2 × 10⁶) × 0.00009

First, convert 0.00009 to scientific notation:

0.00009 = 9 × 10 ⁻⁵

Then

(2 × 10⁶) × 0.00009 = (2 × 10⁶) × (9 × 10 ⁻⁵)

… = (2 × 9) × (10⁶ × 10 ⁻⁵)

… = 18 × 10¹

… = 1.8 × 10²

4 0

2 years ago

I don't understand what to do. can u please answer these questions

MariettaO [177]

I believe you will need to find what -0.125 of an inch is, subtract that from 16, then divide that into 11.

6 0

3 years ago

Which is the graph of the linear inequality y<3x+1

Varvara68 [4.7K]

Answer:

This is in the form y=mx+b . m is the slope and b is the y-intercept. In this case, the slope is 3 and the y-intercept is −1 .

5 0

3 years ago

PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi

Irina18 [472]

Answer:

Question 1: $P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}$

Question 2:

A. $P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

B. $P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

C. $P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

Step-by-step explanation:

Conditional probability is defined by

$P(A|B)= \frac{P(A and B)}{P(B)}$

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} } = \frac{1}{2}$

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

$P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

then

$P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

$P((YorZ) and B)= \frac{3}{16}$

By definition,

$P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

3 0

3 years ago

Given a population mean of 53.7 with a standard deviation of 1.3 and a sample size

a_sh-v [17]

Answer: A. 0.4899 B. 14

Step-by-step explanation:

If X = random variable

then, the standard deviation of the sampling distribution of X = $\dfrac{\sigma}{\sqrt{n}}$

GIven: $\sigma= 1.2, \n= 6$

The standard deviation of the sampling distribution of X = $\dfrac{1.2}{\sqrt{6}}=0.4899$

B. Let n be the sample size,

$0.325=\dfrac{1.2}{\sqrt{n}}\\\\\Rightarrow\ \sqrt{n}=\dfrac{1.2}{0.325}\\\\\Rightarrow\ \sqrt{n}=3.69230769231\\\\\Rightarrow\ n= (3.69230769231)^2=13.6331360947\approx 14$

Hence, the required sample size = 14

3 0

2 years ago