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Elis [28]
3 years ago
11

A Statistics class has a test average of 87 with a standard deviation of 2.1. Find the z-score that corresponds to Kate's test s

core of 93. (Round to the nearest hundredths)
Mathematics
1 answer:
Leto [7]3 years ago
3 0

Answer:

2.86

Step-by-step explanation:

So I'm pretty sure we don't want the probability, just the zscore

to find the zsore subtract the mean and divide by the standard deviation

(93-87)/2.1= 2.86

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28÷7+(-19+10)(-4) <br>does anyone know how to do this <br>​
ikadub [295]
Pemdas
28/7 + (-19+10)(-4)
28/7+ (-9)(-4)
28/7 + (36)
4+ (36)
40
5 0
3 years ago
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
2 years ago
What is the solution to the system of equations y= -3x+6 and y =9
Lilit [14]
Plug the value of 9 in for y
9=-3x+6

Subtract 6 from both sides
3=-3x

Divide both sides by -3
-1=x

Final answer: x=–1
7 0
3 years ago
Somebody please help me with this
kap26 [50]

Answer:

D. $7/hour

Step-by-step explanation:

128-86=42

42/6=7

6 0
3 years ago
Read 2 more answers
Cos s=-2/5 and sin t=4/5, s and t are in quadrant II<br> find cos(s+t) and cos(s-t)
mojhsa [17]

Answer:

•cos(s+t) = cos(s)cos(t) - sin(s)sin(t) = (-⅖).(-⅗) - (√21 /5).(⅘) = +6/25 - 4√21 /25 = (6-4√21)/25

•cos(s-t) = cos(s)cos(t) + sin(s)sin(t) = (-⅖).(-⅗) + (√21 /5).(⅘) = +6/25 + 4√21 /25 = (6+4√21)/25

cos(t) = ±√(1 - sin²(t)) → -√(1 - sin²(t)) = -√(1 - (⅘)²) = -⅗

sin(s) = ±√(1 - cos²(s)) → +√(1- cos²(s)) = +√(1 - (-⅖)²) = √21 /5

5 0
3 years ago
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