All categories

4 years ago

at an amusement park, a person spends $30 on admission and food, and then goes on r number of rides that cost $2 each

1 answer:

3 0

30 dollars would be your initial amount which would be the b part of your equation, y=mx +b. 2r would be the mx part. Your equation would read y=2r+30 and then you would plug in how many rides they went on into the r space to find y

You might be interested in

Pls help me solve this

VARVARA [1.3K]

Answer:

Step-by-step explanation:

5 $\frac{5}{6}$ * 4 $\frac{1}{2}$

=5*6+5/6 * 4*2+1/2

=35/6 * 9/2

=35*9/6*2

=315/12

=105/4

=26 $\frac{1}{4}$

5 0

3 years ago

What us the answer to x + 6 = - 10

Vlad [161]

Answer:

-16

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

An album receives an award when it sells 100,000,000 copies. An album has sold 7,680,000 copies. How many more copies does it ne

Sloan [31]

Answer:

92,320,000

Step-by-step explanation:

100,000,000 - 7,680,000 = 92,320,000

branliest? :)

7 0

3 years ago

What is into relation with suitable example

Goryan [66]

Answer:

the relation between the x-values and y-values of ordered pairs

5 0

3 years ago

There are only 4 mint biscuits and 1 toffee biscuit in a tin. There are only 5 mint sweets and 3 strawberry sweets in a packet.

Debora [2.8K]

Answer:

0.425

Step-by-step explanation:

Given that there are 4 mint biscuits and 1 toffee biscuit in a tin and 5 mint sweets and 3 strawberry sweets in a packet.

So, the total number of biscuit in tin = 4+1= 5 and

Number of mint biscuit in tin = 4

Number of toffee biscuit in tin = 1

So, the probability of picking 1 mint biscuit from tin,

$p_1$ =(Number of mint biscuit in tin)/(Total number of biscuit in the tin)

$\Rightarrow p_1 =4/5 = 0.8$

The probability of picking 1 toffee biscuit from tin,

$p_2$ =(Number of toffee biscuit in tin)/(Total number of biscuit in the tin)

$\Rightarrow p_2=1/5 = 0.2$

Similarly, the total number of sweets in the basket = 5+3=8 and

Number of mint sweets in the basket = 5

Number of strawberry sweets in the basket = 3

So, the probability of picking 1 mint sweets from the basket,

$p_3$ =(Number of mint sweets in the basket)/(Total number of sweets in the basket)

$\Rightarrow p_3=5/8 = 0.625$

The probability of picking 1 strawberry sweets from the basket

$p_4$ =(Number of strawberry sweets in the basket)/(Total number of sweets in the basket)

$\Rightarrow p_4=3/8 = 0.375$

As Michael takes a biscuit from the tin as well as a sweet from the packet randomly, so the possible combinations can be

A) 1 mint biscuit and 1 mint sweets,

B) 1 toffee biscuit and 1 strawberry sweets,

C) 1 mint biscuit and 1 strawberry sweets, and

D) 1 mint sweets and 1 toffee biscuit.

Here, (C) and (D) are the desired case of either the biscuit is mint, but not both.

So, the probability of either the biscuit is mint, but not both

= Probability of ( 1 mint biscuit and 1 strawberry sweets) or (1 mint sweets and 1 toffee biscuit)

$=p_1 \times p_4 + p_2\times p_3 \\\\=0.8 \times 0.375 + 0.2\times 0.625 \\\\=0.425$

Hence, the probability of either the biscuit is mint, but not both is 0.425

4 0

3 years ago