All categories

2 years ago

Three bananas and two pears cost $8. five banana and three pear cost $13

Mathematics

1 answer:

7nadin3 [17]2 years ago

7 0

Answer:

406

Step-by-step explanation:

mnubytvcsdfghcrevthrrtd3s2erecrstvr

You might be interested in

Linda has already written 37 pages and she writes 8 pages per hour. Let P be the total pages written and h the number of hours s

Klio2033 [76]

Answer:

Step-by-step explanation:

7 0

2 years ago

Please help me<br><br> Find the values of x and y

MaRussiya [10]

(12x +15)°=75° .…(शिर्षभिमुख कोण भएर)

or, 12x+15=75

or, 12x=75-15

or, 12x=60

or, x=60÷12

or, x=5

(6y-27)°=105° .…(शिर्षभिमुख कोण भएर)

or, 6y-27=105

or, 6y=105+27

or, 6y=132

or, y=132÷6

or, y=22

5 0

1 year ago

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r

kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

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.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that $p = 0.15$

Assume you obtain a random sample of 9 individuals from this population:

This means that $n = 9$

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

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

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

$P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005$

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0

2 years ago

The accompanying line chart shows the major spending categories of the federal budget over the last 50 years. (Payments to indi

Galina-37 [17]

Answer:

idk but i think its -5%

Step-by-step explanation:

3 0

3 years ago

A truck company charges $30 to rent a moving truck for the day, plus 5 cents for every mile driven during the day. The function

statuscvo [17]

Answer:

I just finish my Algebra hw to see this??

Step-by-step explanation:

6 0

2 years ago