I really need help ................................................

a boy sold his bike and accessories for $168 if he received 3 times as much for the bike as for the accessories how much did he

Veronika [31]

168/4 to get the bike and the accessories to get 42
42x3=126

3 0

3 years ago

The length of a rectangle is 4 more than 3 times the width if the perimeter of the rectangle is 18.4 cm what is the area of the

garik1379 [7]

Let's call the width: w

the lenght is then 3w+4 ("4 more than 3 times the width")

and the parameter would be 2(w+3w+4), that is 2*(4w+4), that is 8w+8.

this is also equal to 18.4:

8w+8=18.4
8w=10.4
w=1.3

this is the width, and the lenght is:
4+3*1,3=4+3.9=7.9

and the area is their product:
1.3*7.9=10.27

6 0

3 years ago

The Perimeter is x+38

OLga [1]

P=x+38

add all individual sides together (is also the perimeter):
p=(x+3)+(x-5)+x+(x-2)+(x-4)+(x-1)+(x-1)
=7x-10

set both p equal:
x+38=7x-10
38=6x-10
48=6x
8=x

then insert x=8 in all the side/perimeter equations:
p=x+38=8+38=46

sides:
x=8
x-2=8-2=6
x-1=8-1=7
x-4=8-4=4
x-5=8-5=3
x+3=8+3=11

8 0

3 years ago

Suppose that out of 20% of all packages from Amazon are delivered by UPS. 15% of the packages that are delivered by UPS weighs 2

Gelneren [198K]

Answer:

0.1938 = 19.38% probability that a package is delivered by UPS if it weighs 2 lbs or more

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: It weighs 2 lbs or more

Event B: Delivered by UPS

15% of the packages that are delivered by UPS weighs 2 lbs or more.

This means that $P(A \cap B) = 0.15$

More than 2 lbs:

15% of 20%(UPS)

93% of 80%(Not delivered by UPS). So

$P(A) = 0.15*0.2 + 0.93*0.8 = 0.774$

a. What is the probability that a package is delivered by UPS if it weighs 2 lbs or more

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.774} = 0.1938$

0.1938 = 19.38% probability that a package is delivered by UPS if it weighs 2 lbs or more

3 0

2 years ago

a product code consists of picking a two-digit number (including 00) and 3 not necessarily distinct letters. How many codes are

Brrunno [24]

Answer:

The codes are:

N,_,_,_

Where each _ represents a possible letter, out of 26.

And N is a number of two digits.

So in total we can think that we have 5 empty slots.

_,_,_,_,_

In the first slot we can put any digit between 0 and 9, so here we have 10 options.

In the second slot we can put any digit between 0 and 9, so here we have 10 options.

In the third slot we can put any letter, so here we have 26 options (and exactly the same for the fourth and fifth slots)

The total number of different combinations is equal to the product of the number of options for each slot.

C = 10*10*26*26*26 = 1,757,600

Now, if we want to have at least one nine, we can fix it in the first slot.

Then we have:

in the first slot we have only one option (the 9)

In the second slot we can put any digit between 0 and 9, so here we have 10 options.

In the third slot we can put any letter, so here we have 26 options (and exactly the same for the fourth and fifth slots)

The number of combinations is:

C = 1*10*26*26*26 = 175,760

But we also should consider the case where we fix the 9 in the second slot, so the actual number of combinations is twice the number above.

C = 2*175,760 = 351,520

The probability that the code does not contain the number 9.

Now in the first slot we have only 9 options, all the whole numbers between 0 and 8.

The same for the second slot, 9 options.

For the third, fourth and fifth slot is the same as before.

The total number of combinations is:

C = 9*9*26*26*26 = 1,423,656

4 0

3 years ago

I really need help ................................................​

I really need help ................................................