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3 years ago

Yall i need this to be corret please explain giving brainliest please

Mathematics

2 answers:

Deffense [45]3 years ago

6 0

Answer:

A) both expression involve addition

ankoles [38]3 years ago

4 0

The answer is C I am very positive each expression includes the term x

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So, I need to find the Mean Median Mode and Range with this frequency table, and I have no idea how to do that. Can somebody ple

olya-2409 [2.1K]

Mean: average of the numbers. Add them up, divide by how many numbers/entries there are.
-2 + -1 + 0 + 0 + 0 + 0 + 2 + 4 = 3
3 divided by 8 = 0.375
Your mean is 0.375

Median: write the data in numerical order, find the middle number.

-2, -1, 0, 0, 0, 0, 2, 4
In this data set, there are an even number of entries, so we average the middle two numbers. Thankfully, here, the middle two are the same, so your median is 0.

Mode: the number that appears the most often in the data set

Your mode is also 0, because there are more zeroes than any other number in the data set.

Range: the distance on a number line between the highest and lowest number.

The distance between -2 and 4 is 6.4 - (-2) = 6

Please LMK if you have questions

8 0

3 years ago

25 Points! Midterm Today! Can anyone help with this problem?

Andrews [41]

I really do wish I could help But I’m not sure I tried to figure it out and i just can’t get it

6 0

3 years ago

Image attached: some sort of triangle stuff

Drupady [299]

Answer:

Step-by-step explanation:

On edg 2020

8 0

3 years ago

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used 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.

41% of U.S. adults have very little confidence in newspapers.

This means that $p = 0.41$

You randomly select 10 U.S. adults.

This means that $n = 10$

(a) exactly five

This is $P(X = 5)$ . So

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

$P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087$

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480$

$P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125$

$P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019$

$P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001$

Then

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834$

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

(c) less than four.

This is:

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

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

$P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051$

$P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355$

$P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111$

$P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575$

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0

3 years ago

In rectangle MNPQ with side MN = 16 in, point L is the midpoint of PQ and m∠QLM = 45°. Find the perimeter of the rectangle.

Marina86 [1]

See the attached figure

we know that
MN=16 in-------------> PQ=MN
PQ=16 in
<span>L is the midpoint of PQ-----------> distance QL=</span><span>PQ/2
QL=8 in
if </span>m∠QLM = 45°
then
MQ=QL
MQ=8 in

Perimeter=[MN]*2+[MQ]*2=16*2+8*2=48 in

the answer is 48 in

7 0

3 years ago