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2 years ago

15

A meteorologist reported that in the month of April there were 3 inches more rainfall than normal. Write an integer to represent

the amount of rainfall above normal in April.

2 answers:

BartSMP [9]2 years ago

8 0

Answer:

3

Step-by-step explanation:

mel-nik [20]2 years ago

7 0

Answer:3

Step-by-step explanation:my teacher told me hope it correct it was correct for me

You might be interested in

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

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.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

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

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

2 years ago

A farmer has 520 feet of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one s

Setler [38]

Answer:

$310\text{ feet and }210\text{ feet}$

Step-by-step explanation:

GIVEN: A farmer has $520 \text{ feet}$ of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one side of the pen. The length of the barn is $310 \text{ feet}$ .

TO FIND: Determine the dimensions of the rectangle of maximum area that can be enclosed under these conditions.

SOLUTION:

Let the length of rectangle be $x$ and $y$

perimeter of rectangular pen $=2(x+y)=520\text{ feet}$

$x+y=260$

$y=260-x$

area of rectangular pen $=\text{length}\times\text{width}$

$=xy$

putting value of $y$

$=x(260-x)$

$=260x-x^2$

to maximize $\frac{d \text{(area)}}{dx}=0$

$260-2x=0$

$x=130\text{ feet}$

$y=390\text{ feet}$

but the dimensions must be lesser or equal to than that of barn.

therefore maximum length rectangular pen $=310\text{ feet}$

width of rectangular pen $=210\text{ feet}$

Maximum area of rectangular pen $=310\times210=65100\text{ feet}^2$

Hence maximum area of rectangular pen is $65100\text{ feet}^2$ and dimensions are $310\text{ feet and }210\text{ feet}$

5 0

2 years ago

IT IS DUE AT 11:59PM. PLEASE HELP!!!!

kaheart [24]

Hello from MrBillDoesMath!

Answer: b = 4, h = 13

Discussion:

For a triangle with base b" and height "h, the area formula is

A= (1/2)bh.

In our case A = 26, h = 5b-7, so

26 = (1/2) b (5b-7). Multiply both sides by 2:

26*2 = b(5b-7) =>

52 = b(5b-7) =>

52 = 5b^2 - 7b => (subtract 52 from both sides)

5b^2 - 7b - 52 = 0. This factors as follows:

(b - 4) (5 b + 13) = 0 (use quadratic formula to find this)

so b = 4 and h = 5b -7 = 5*4 - 7 = 13

Thank you,

MrB

3 0

2 years ago

Read 2 more answers

A<br><br> B<br><br> C<br><br> Neither of them

lions [1.4K]

None of them are correct.

8 0

2 years ago

Your carry-on bag can weigh at most 40 pounds.Select an inequality that represents how much more weight x (in pounds) you can ad

andrey2020 [161]

Answer:

22 + x ≤ 40

x ≤ 18 pounds

Step-by-step explanation:

Maximum weight = 40 pounds

Initial weight = 22 pounds

Let Maximum weight that can be added = x

Mathematically, this means :

Initial weight + additional weight ≤ 40

That is ;

22 + x ≤ 40

Additional weight, x :

x ≤ 40 - 22

x ≤ 18 pounds

7 0

2 years ago