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3 years ago

What is the constant of proportionality in the proportional relationship y= 4/5x ?

Mathematics

2 answers:

Lady bird [3.3K]3 years ago

7 0

Answer:

the snserw should be -3 or -5

Step-by-step explanation:

SCORPION-xisa [38]3 years ago

3 0

Answer:

Step-by-step explanation:

Proportional relationship is:

y = kx, where k is the constant of proportionality

We have:

y = 4/5x

So the value of k is 4/5

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2x-5y=10 In steps pls?

nataly862011 [7]

Answer:

Good bye . Please MARK me the BRAINLIEST

8 0

3 years ago

Which best describes the composition of tranformations that maps ∆LMN to ∆L'M'N'. 0,0 is the center of the dilation in each choi

Eduardwww [97]

Triangle LMN was dilated by a scale factor of 2 followed by a translation 1 unit right and 3 unit up to form triangle L'M'N'

<h3>What is transformation?</h3>

Transformation is the movement of a point from its initial location to a new location. Types of transformations are reflection, translation, rotation and dilation.

Rigid transformation preserves the shape and size of the figure. Reflection, translation, rotation are rigid transformations.

Triangle LMN was dilated by a scale factor of 2 followed by a translation 1 unit right and 3 unit up to form triangle L'M'N'

Find out more on transformation at: brainly.com/question/4289712

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7 0

1 year ago

Jenna has a piece of paper that is 84 square inches in area. it is 10 1/2 inches long. what is the length of the piece of paper

Ronch [10]

I hope this helps you

6 0

3 years ago

True or False:

atroni [7]

Answer:

no it can't because

Step-by-step explanation:

coz the sum of any two angle should be greater than 3rd side

13 + 5 = 18 which is greater than 7
13 + 7 = 20 which is greater than 5
5 + 7 = 12 which is not greater than 13

so triangle cant be made

3 0

3 years ago

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development

Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

5 0

3 years ago