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2 years ago

HELPPPPPPPPPPP APPARENTLY MY ANSWER WAS WRONG

Mathematics

2 answers:

Nonamiya [84]2 years ago

8 0

Answer:

d= -12 + 925

Step-by-step explanation:

decrease means negative and Increase means up (positive) so probably that why your answer was wrong

densk [106]2 years ago

8 0

Answer:

d= -12 + 925

Step-by-step explanation:

You might be interested in

What is the length of AB?

makkiz [27]

The radius of the circle is given to be r = 14.5

Let the center be O. So OB is the radius = 14.5

We can draw a triangle as shown in the image below. We have a right angled triangle. We know the hypotenuse and the base, we need to find the perpendicular side. This can be done using the Pythagorean theorem.

So, we can write:

$(14.5)^{2}=10^{2}+OA^{2} \\ \\ 
OA^{2}=110.25 \\ \\ 
OA=10.5$

Thus, the measure of OA = 10.5

OB = OA + AB

14.5 = 10.5 + AB
⇒
AB = 4

3 0

2 years ago

Read 2 more answers

53% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

faltersainse [42]

Answer:

$P(X = 5) = 0.2423$

$P( X \ge 6 )= 0.4516$

$P( X < 4 )= 0.12694$

Step-by-step explanation:

From the question we are told that

The population proportion is p = 0.53

The sample size is n = 10

Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution

i.e

$X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial distribution is

$P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

Here C stands for combination hence we are going to be making use of the combination function in our calculators

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly five is mathematically represented as

$P(X = 5) = ^{10}C_5 * 0.53^5 * (1- 0.53)^{10-5}$

=> $P(X = 5) = 252 * 0.04182 * 0.023$

=> $P(X = 5) = 0.2423$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is at least six is mathematically represented as

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

=> $P( X \ge 6 )= [^{10}C_6 * [0.53]^6 * (1- 0.53)^{10-6}] + [^{10}C_7 * [0.53]^7 * (1- 0.53)^{10-7}] + [^{10}C_8 * [0.53]^8 * (1- 0.53)^{10-8}] + [^{10}C_9 * [0.53]^9 * (1- 0.53)^{10-9}] + [^{10}C_{10} * [0.53]^{10} * (1- 0.53)^{10-10}]$

=> $P( X \ge 6 )= [0.227] + [0.1464] + [0.0619] + [0.0155] + [0.00082]$

=> $P( X \ge 6 )= 0.4516$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is less than four is mathematically represented as

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

=> $P( X < 4 )= [^{10}C_3 * 0.53^3 * (1- 0.53)^{10-3}] + [^{10}C_2 * 0.53^2 * (1- 0.53)^{10-2}] + [^{10}C_1 * 0.53^1 * (1- 0.53)^{10-1}] + [^{10}C_0 * 0.53^0 * (1- 0.53)^{10-0}]$