All categories

2 years ago

PLEASE HELP NOW!!!

Mathematics

1 answer:

jeka942 years ago

7 0

Step-by-step explanation:

part A

c=3.14d

3.14*8

=25.12inches

circumference circle B

3.14*3

=9.42inches

part B area

A=3.14r^2

=3.14*4*4

=50.24inches^2

=3.14*1.5*1.5

=7.065inches^2

part c

the circumference area are larger for circles with bigger diameter or radius.

You might be interested in

50 POINTS. PLEASE HELP!!! NEED THIS DINE BY 4:00 PM!!!!

malfutka [58]

Hey fellow Army.

So because JKLM=PQRS, that means that all the line segments are equal.

JK=PQ=6.4

KL=QR=13

LM=RS=9

MJ=SP=18

Okay now for the graph.

It was first reflected over the y axis, and then shifted over vertically by 9 .

Hope this helps. Have a good day!

4 0

3 years ago

Read 2 more answers

Select the best answer for the question.

LiRa [457]

Answer: the answer is B

Step-by-step explanation:

8 0

3 years ago

Work out the area of a parallelogram with a base of 6cm and height 8cm

FromTheMoon [43]

Multiply 6 and 8 which is 48cm

5 0

3 years ago

Read 2 more answers

Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list

Radda [10]

Answer:

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: Any from 0 to 5.

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

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 question:

$n = 5, p = 0.3, q = 1 - p = 0.7$

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

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

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

8 0

4 years ago

Answer fraction in lowest terms

Marat540 [252]

7/16 should be the answer

6 0

3 years ago

Read 2 more answers