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3 years ago

Josefina origanally had $550 in her checking account. After withdrawing 100$ each week, Josefina has a new balance of $50.

Mathematics

2 answers:

Amanda [17]3 years ago

8 0

W=5. She withdrew money 5 times.

ddd [48]3 years ago

6 0

$5beacuse u need to subtract

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Which set of numbers could represent the lengths of the sides of a right triangle?

DiKsa [7]

Answer:

The first set: 8, 15, and 17.

Step-by-step explanation:

By the pythagorean theorem, a triangle is a right triangle if and only if

$\text{longest side}^2 = \text{first shorter side}^2 + \text{second shorter side}^2$ .

In this case,

$\text{longest side}^2 = 17^2 = 289$ .

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 8^2 + 15^2\\ &=64 + 225 = 289 \end{aligned}$ .

In other words, indeed $\text{hypotenuse}^2 = \text{first leg}^2 + \text{second leg}^2$ . Hence, 8, 15, 17 does form a right triangle.

Similarly, check the other pairs. Keep in mind that the square of the longest side should be equal to the sum of the square of the two

Factor out the common factor $2$ to simplify the calculations.

$\text{longest side}^2 = 20^2 = 400$

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 10^2 + 15^2\\ &=100 + 225 = 325 \end{aligned}$ .

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

$\text{longest side}^2 = (2\times 11)^2 = 2^2 \times 121$ .

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= (2 \times 6)^2 + (2 \times 9)^2\\ &=2^2 \times(36 + 81) = 2^2 \times 117 \end{aligned}$ .

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

$\text{longest side}^2 = 11^2 = 121$ .

<h3> $\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 7^2 + 9^2\\ &=49+ 81 = 130 \end{aligned}$ .</h3>

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

6 0

3 years ago

a package of marbles contains 20 blue marbles, 38 red marbles, 48 green marbles, and 16 yellow marbles. which color of marbles c

OLEGan [10]

the answer would be 48 green marbles

Step-by-step explanation:

48/6=8

So, each person would get 8 marbles each

7 0

2 years ago

Your school's football team scored 49 points. Your team's score was 19 points more than the opponent's score s. Write and solve

solmaris [256]

Answer:

39 points

Step-by-step explanation:

Your school's football team scored 49 points.

Your team's score is 19 points more than the opponents score

The opponent score is represented as S

Since the total points that was scored is 49 and your team score is 19 then the equation can be written as follows

S+ 19= 49

Solve for S

S= 49-19

S= 30

Hence the opponent's score is 30 points

7 0

3 years ago

A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o

o-na [289]

Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that $\mu = \frac{n}{60}$ , in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

$n = 25$ , and thus, $\mu = \frac{25}{60} = 0.4167$

This probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408$

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that $n = 6$

Each of these days, 0.6592 probability of no accident on this stretch, which means that $p = 0.6592$ .

This probability is P(X = 2). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879$

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

$\mu = 0.6\frac{n}{60} = 0.01n$