A coin is tossed four times. What is the probability of getting tails four times in a row?

A math test is worth 100 points and has 38 problems. each problem is worth either 5 points or 2 points. how many problems of eac

blondinia [14]

Hey there,
There are 2 ways
1st way:
Since there are 8 5 point problems, 38 - 8 = 30, will give you the 2 point problems.

2nd way:
8 x 5 = 40
100 - 40 = 60
60 / 2 = 30
Thus there are 30, 2 point problems.

Hope this helps :))

~Top

4 0

3 years ago

Read 2 more answers

Hi please help thank you i’ll give brainliest

Sergio [31]

Answer:

ok so first start with 97 and 45 put together 57/97 and

Step-by-step explanation:

4 0

3 years ago

For each equation, determine whether it represents a direct variation, an inverse variation, or neither.

SSSSS [86.1K]

we know that in direct variation as x increase y also increases.

in indirect variation x decreases y increase and vice versa.

in part a we have x at top (numerator ) and y at denominator it mean indirect variation (as x increases y decreases).

to find k we know that for indirect variation xy=k

if we rewrite the equation x=9/y we get xy=9

which mean k=9 .

in part b we have 4xy=20 if we simplify this equation we get

$xy=20/4=5$

so here if we rewrite in terms of x.

we get x=5/y which represents indirect variation.

and we know xy=k

we have xy=5 .so k=5

8 0

3 years ago

Show solutions please

slega [8]

Answer:

1. Their ages are:

Steve's age = 18

Anne's age = 8

2. Their ages are:

Max's age = 17

Bert's age = 11

3. Their ages are:

Sury's age = 19

Billy's age = 9

4. Their ages are:

The man's age = 30

His son's age = 10

Step-by-step explanation:

1. We make the assumption that:

S = Steve's age

A = Anne's age

In four years, we are going to have:

S + 4 = (A + 4)2 - 2 = 2A + 8 - 2

S + 4 = 2A + 6 .................. (1)

Three years ago, we had:

S - 3 = (A - 3)3

S - 3 = 3A - 9

S = 3A - 9 + 3

S = 3A – 6 …………. (2)

Substitute S from (2) into (1) and solve for A, we have:

3A – 6 + 4 = 2A + 6

3A – 2A = 6 + 6 – 4

A = 8

Substitute A = 8 into (3), we have:

S = (3 * 8) – 6 = 24 – 6

S = 18

Therefore, we have:

Steve's age = 18

Anne's age = 8.

2. We make the assumption that:

M = Max's age

B = Bert's age

Five years ago, we had:

M - 5 = (B - 5)2

M - 5 = 2B - 10 .......................... (3)

A year from now, it will be:

(M + 1) + (B + 1) = 30

M + 1 + B + 1 = 30

M + B + 2 = 30

M = 30 – 2 – B

M = 28 – B …………………… (4)

Substitute M from (4) into (3) and solve for B, we have:

28 – B – 5 = 2B – 10

28 – 5 + 10 = 2B + B

33 = 3B

B = 33 / 3

B = 11

If we substitute B = 11 into equation (4), we will have:

M = 28 – 11

M = 17

Therefore, their ages are:

Max's age = 17

Bert's age = 11.

3. We make the assumption that:

S = Sury's age

B = Billy's age

Now, we have:

S = B + 10 ................................ (5)

Next year, it will be:

S + 1 = (B + 1)2

S + 1 = 2B + 2 .......................... (6)

Substituting S from equation (5) into equation (6) and solve for B, we will have:

B + 10 + 1 = 2B + 2

10 + 1 – 2 = 2B – B

B = 9

Substituting B = 9 into equation (5), we have:

S = 9 + 10

S = 19

Therefore, their ages are:

Sury's age = 19

Billy's age = 9.

4. We make the assumption that:

M = The man's age

S = His son's age

Therefore, now, we have:

M = 3S ................................... (7)

Five years ago, we had:

M - 5 = (S - 5)5

M - 5 = 5S - 25 ................ (8)

Substituting M = 3S from (7) into (8) and solve for S, we have:

3S - 5 = 5S – 25

3S – 5S = - 25 + 5

-2S = - 20

S = -20 / -2

S = 10

Substituting S = 10 into equation (7), we have:

M = 3 * 10 = 30

Therefore, their ages are:

The man's age = 30

His son's age = 10

3 0

3 years ago

1.) What is the equation of the path of firework #1? Write your equation in general form.

valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in $y=ax^2+bx+c$

it's easier to find vertex form first then expand to get general form

vertex form is $y=a(x-h)^2+k$ where the vertex is (h,k) and a is a constant

firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

$h_1=a(t-10)^2+50$

to find the value of $a$ , subsitute another point

(0,0)

$0=a(0-10)^2+50)$

$0=100a+50$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_1=\frac{-1}{2}(t-10)^2+50$

expand to get general form

$h_1=\frac{-1}{2}(t^2-20t+100)+50$

$h_1=\frac{-1}{2}t^2+10t-50+50$

$h_1=\frac{-1}{2}t^2+10t$

2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is $h_2=a(t-10)^2+72$

find $a$

use another point

(0,22)

$22=a(0-10)^2+72$

$22=100a+72$

$-50=100a$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_2=\frac{-1}{2}(t-10)^2+72$

3.

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50

domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

3 0

3 years ago