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3 years ago

9

The scatter plot below shows the ages and heights of a varsity basketball team. Each dot represents one player. What is the tota

l number of 17-year olds whose height is 74 inches or less?
A. 0
B. 1
C. 2
D. 3

2 answers:

sleet_krkn [62]3 years ago

3 0

Answer:

1

Step-by-step explanation:

krok68 [10]3 years ago

3 0

Answer:

C. 2

Step-by-step explanation:

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In ΔHIJ, the measure of ∠J=90°, IH = 5, HJ = 4, and JI = 3. What is the value of the tangent of ∠I to the nearest hundredth?

salantis [7]

Answer= 4/3 or 1.33

Explanation:

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2 years ago

If 1/2x+1/3=4 What is the value of 3x+2y

ivolga24 [154]

Answer:

x=3/22 ; y=-9/44

Step-by-step explanation:

1/2x+1/3=4

=>1/2x=4-1/3

=>1/2x=(12-1)/3

=>1/2x=11/3

=>22x=3

=>x=3/22....(1)

then,

3x+2y

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3 years ago

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Anna11 [10]

Answer:

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Step-by-step explanation:

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2 years ago

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Verify: sec^2x+tan^2x=2sec^2x-1

Jlenok [28]

Answer:

see explanation

Step-by-step explanation:

Using the identity

tan²x = sec²x - 1

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2 years ago

An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

$P(X = x) = \frac{y^xe^{-y}}{x!}$

Where y represents λ (average)

y = 2.3

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

Using probability notations;

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

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

$P(X = 0) = \frac{1 * e^{-2.3}}{1}$

$P(X = 0) = e^{-2.3}$

$P(X = 0) = 0.10025884372$

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

$P(X = 1) =2.3 * e^{-2.3}$

$P(X = 1) = 2.3 * 0.10025884372$

$P(X = 1) = 0.23059534055$

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

$P(X\leq 1) = 0.10025884372 + 0.23059534055$

$P(X\leq 1) = 0.33085418427$

$P(X\leq 1) = 0.331$

Hence, the required probability is 0.331

6 0

2 years ago