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2 years ago

For each sequence below, identify the next two terms. 5, 1, 7, 3, 9... *

Mathematics

1 answer:

Paul [167]2 years ago

8 0

Answer:

5, 11

Step-by-step explanation:

the number is subtracting 4 adding 6 subtracting 4 adding 6

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You received partial credit in th

Blababa [14]

Answer:

Expected value would be $ 0.896

Step-by-step explanation:

Given,

The price of the lottery ticket = $44800000,

Also, the probability of winning the grand prize = .000000020,

Thus, the expected value of the lottery ticket = value of the lottery ticket × probability of getting the lottery ticket

= 44800000 × .000000020

= $0.896

Note : value of lottery ticket = prize amount - cost of each ticket,

Here the cost price of a ticket is not given,

That's why we did not consider it.

5 0

3 years ago

PLEASEEEEE SOMEBODY HELP NOW QUICK PLEASEEE

pickupchik [31]

Answer:

Step-by-step explanation:

becasue it would have a 3/8 chance of landing on silver , 8 goes into 240 30 times. so if we multiply 3 by 30 it gives us 90.

3 0

2 years ago

Thor invests $3000 in an account, with interest compounded continuously. If his investment doubles in value after 7.2 years, wha

sladkih [1.3K]

Answer:

B. 9.6 %.

Step-by-step explanation:

A = P e^rt

His investment doubles so it comes to 3000 * 2 = $6000.

Substituting in the formula>

6000 = 3000e^7.2r ehere r is the interest rate.

e^7.2r = 2

Taking logs

7.2r = ln 2

r = ln2 / 7.2

r = 0.096

So the rate is 9.6%.

8 0

2 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

Please help me with problem 1 of Properties of Multiplication with Whole Numbers

kobusy [5.1K]

This is the associative property, since any numbers can <em>associate </em>with each other, via the parentheses.

8 0

3 years ago