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beks73 [17]
3 years ago
13

Please anwser fast it’s due in 5 minutes

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
6 0

Answer:30000 just 5000+  six time

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Which measures of central tendency would be affected if the outlier 11 was added to the following set 27,20,34,37,21,42,39
MrRa [10]

Answer:

Median will not be affected by the outlier.

Step-by-step explanation:

With the outlier, the mean will be dragged way down. The median will likely be about the same. Mean is non-resistant to outliers, median is resistant.

Hope this helps!

5 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
olga nikolaevna [1]
<h3>Answer:   1.15</h3>

==============================================================

Work Shown:

QR = 73

QP = 55

PR = x

Pythagorean Theorem

a^2 + b^2 = c^2

(QP)^2 + (PR)^2 = (QR)^2

(55)^2 + (x)^2 = (73)^2

3025 + x^2 = 5329

x^2 = 5329-3025

x^2 = 2304

x = sqrt(2304)

x = 48

PR = 48

----------

tan(angle) = opposite/adjacent

tan(R) = QP/PR

tan(R) = 55/48

tan(R) = 1.1458333 approximately

tan(R) = 1.15

8 0
3 years ago
Read 2 more answers
(a) A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure
Temka [501]

Answer:

For first lamp ; The resultant probability is 0.703

For both lamps; The resultant probability is 0.3614

Step-by-step explanation:

Let X be the lifetime hours of two bulbs

X∼exp(1/1400)

f(x)=1/1400e−1/1400x

P(X<x)=1−e−1/1400x

X∼exp⁡(1/1400)

f(x)=1/1400 e−1/1400x

P(X<x)=1−e−1/1400x

The probability that both of the lamp bulbs fail within 1700 hours is calculated below,

P(X≤1700)=1−e−1/1400×1700

=1−e−1.21=0.703

The resultant probability is 0.703

Let Y be a lifetime of another lamp two bulbs

Then the Z = X + Y will follow gamma distribution that is,

X+Y=Z∼gamma(2,1/1400)

2λZ∼

X+Y=Z∼gamma(2,1/1400)

2λZ∼χ2α2

The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,

P(Z≤1700)=P(1/700Z≤1.67)=

P(χ24≤1.67)=0.3614

The resultant probability is 0.3614

8 0
3 years ago
How do you prove that a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects?
konstantin123 [22]
Check the picture drawn

Consider the line segment AB, 

let M be the midpoint of AB, so AM=MA

erect the perpendicular MT to AB from point M. Pick a point P on MT and join it to the points A and B

The triangles PMA and PMB are congruent from the Side Angle Side congruence postulate:

AM=MA, PM is common and m(PMA)=m(PMB)=90°, as MT is perpendicular to AB

so PA=PB

5 0
3 years ago
What is the surface area of a box with a length of 6 ft, a width of 3 ft, and a height of 2½ ft?
NikAS [45]

Answer:

Step-by-step explanation:

Remark

The surface area is the area of all six sides when the object is a box.

Formula

SA = 2*L*w + 2*L*h + 2*w*h

Givens

L = 6 ft

w= 3 ft

h = 2.5 feet

Substitute and Solution

SA = 2*6*3 + 2*6*2.5 + 2*3*2.5

SA = 36 + 30 + 15

SA = 81

4 0
2 years ago
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