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3 years ago

Math and answer please than you

Mathematics

2 answers:

finlep [7]3 years ago

6 0

Answer:

Step-by-step explanation:

C^2 = A^2 + B^2

plug in numbers

100 = 64 + B^2

36 = b^2

square

b = 6

elena55 [62]3 years ago

5 0

Answer:

Step-by-step explanation:

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How do you write 1.7 in expanded form

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1.7 = (1 x 1) + (7/10)

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3 years ago

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#1 Identify the characteristics of the quadratic graph

shusha [124]

Vertex: (1,-3)
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Find the constant of variation (k) where y varies directly with x. y = 2, when X = 6.

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Answer:

The graph below models the value of a $20,000 car t years after it was purchased.

Value of Car

A graph titled Value of Car has years on the x-axis and Dollars on the y-axis. A line curves down and goes through points (0, 20,000), (4, 10,000), and (14, 2,000).

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Each time, t, is associated with exactly one car value, y.

The rate at which the car decreases in value is not constant.

There is no time, t, at which the value of the car is 0.

Step-by-step explanation:

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3 years ago

two cards are selected from a standard deck of 52 playing cards. the first card is not replaced before the second card is select

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3 years ago

An analysis of sales records for the last 120 weeks gives the following results. Assuming that these past data are a reliable gu

Mila [183]

Answer:

(a) 0.5333

(b) 0.6583

(d) 0.7083

(e) 0.6167

Step-by-step explanation:

Denote the events as follows:

A = a competitor will advertise

NA = a competitor will not advertise

L = Low sales will be achieved

M = Medium sales will be achieved

H = High sales will be achieved

The data provided is of the form:

Low (L) Medium (M) High (H) Total

A 32 14 18 64

NA 21 12 23 56

Total 53 26 41 120

The probability of an event E is:

$P(E)=\frac{n(E)}{N}$

n (E) = favorable outcomes of event E

N = Total number of outcomes

(a)

Compute the probability that next week the competitor will advertise as follows:

$P(A)=\frac{n(A)}{N}=\frac{64}{120}=0.5333$

Thus, the probability that next week the competitor will advertise is 0.5333.

(b)

Compute the probability that next week sales will not be high as follows:

$P(H^{c})=1-P(H)=1-\frac{n(H)}{N}=1-\frac{41}{120}=\frac{120-41}{120}=0.6583$

Thus, the probability that next week sales will not be high is 0.6583.

(c)

The events of achieving a medium or high sales are mutually exclusive.

Since the sales achieved will either be medium or high. They cannot be both.

So, P (M ∩ H) = 0.

Compute the probability that next week there will be medium or high sales will be achieved as follows:

$P(M\cup H)=P(M)+P(H)=\frac{26}{120}+\frac{41}{120}=\frac{26+41}{120}=0.5583$

Thus, the probability that next week there will be medium or high sales will be achieved is 0.5583.

(d)

Compute the probability that next week either the competitor will advertise, or only low sales will be achieved as follows:

$P(A\cup L)=P(A)+P(L)-P(A\cap L)=\frac{64}{120}+\frac{53}{120}-\frac{32}{120}=\frac{85}{120}=0.7083$

Thus, the the probability that next week either the competitor will advertise, or only low sales will be achieved is 0.7083.

(e)

Compute the probability that next week either the competitor will not advertise, or high sales will be achieved as follows:

$P(NA\cup H)=P(NA)+P(H)-P(NA\cap H)=\frac{56}{120}+\frac{41}{120}-\frac{23}{120}=\frac{74}{120}=0.6167$

Thus, the the probability that next week either the competitor will not advertise, or high sales will be achieved is 0.6167.

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4 years ago