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arlik [135]
3 years ago
12

2c+17.6 =6 / 4 SOLVE give your answer as a decimal get brainly

Mathematics
2 answers:
tamaranim1 [39]3 years ago
8 0

Answer:

-8.05

Step-by-step explanation:

In order to solve for C you first have to subtract 17.6 from both sides.

2c=(6/4)-17.6

6/4 is 1.5

1.5-17.6=-16.1

2c=-16.1

c=-8.05

MA_775_DIABLO [31]3 years ago
5 0

Answer:

1.5

Step-by-step explanation:

2c+17.6=6/4

2c+176/10=6/4

2c+88/5=6/4

2c=6/4-88/5

2c=30-352/20

2c=-322/20

2c=-161/10

c=-8.05

Proof:

2c+17.6=6/4

2(-8.05)+17.6=6/4

-16.1+17.6=6/4

1.5=1.5

Hope this helps ;) ❤❤❤

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Rob Herndon, an accountant with Southwest Airlines, wants to retire 50% of
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To achieve his goal of paying off $430,000 in 10 years, Rob Herndon needs to make $34,186.97 at 5% compounded annually.

<h3>What is a future value?</h3>

A future value refers to a value on a future date based on an assumed growth rate of interest over time.

The future value can be determined using the FV formula or table.

We can also compute the future value using an online finance calculator as follows:

<h3>Data and Calculations:</h3>

N (# of periods) = 10 years

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7 0
2 years ago
(43 points) In the US, 85% of the population has Rh positive blood. Suppose we take a random sample of 6 persons and let Y denot
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Answer:

a) Binomial distribution with parameters p=0.85 q=0.15 n=6

b) 62.29%

c) 2.38%

d) See explanation below

Step-by-step explanation:

a)

We could model this situation with a binomial distribution

P(6;k)=\binom{6}{k}p^kq^{6-k}

where P(6;k) is the probability of finding exactly k people out of 6 with Rh positive, p is the probability of finding one person with Rh positive and q=(1-p) the probability of finding a person with no Rh.

So

\bf P(Y=k)=\binom{6}{k}(0.85)^k(0.15)^{6-k}

b)  

The probability that Y is less than 6 is

P(Y=0)+P(Y=1)+...+P(Y=5)

Let's compute each of these terms

P(Y=0)=P(6;0)=\binom{6}{0}(0.85)^0(0.15)^{6}=1.139*10^{-5}

P(Y=1)=P(6;1)=\binom{6}{1}(0.85)^1(0.15)^{5}=0.0000387281

P(Y=2)=P(6;2)=\binom{6}{2}(0.85)^2(0.15)^{4}=0.005486484

P(Y=3)=P(6;3)=\binom{6}{3}(0.85)^3(0.15)^{3}=0.041453438

P(Y=4)=P(6;4)=\binom{6}{4}(0.85)^4(0.15)^{2}=0.176177109

P(Y=5)=P(6;5)=\binom{6}{5}(0.85)^5(0.15)^{1}=0.399334781

and adding up these values we have that the probability that Y is less than 6 is

\sum_{i=1}^{5}P(Y=i)=0.622850484\approx 0.6229=62.29\%

c)

In this case is a binomial distribution with n=200 instead of 6.

p and q remain the same.

The mean of this sample would be 85% of 200 = 170.  

In a binomial distribution, the standard deviation is  

s = \sqrt{npq}

In this case  

\sqrt{200(0.85)(0.15)}=5.05

<em>Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05</em>

So, the approximate probability that there are fewer than 160 persons with Rh positive blood in a sample of 200 would be the area under the normal curve to the left of 160

(see picture attached)

We can compute that area with a computer and find it is  

0.0238 or 2.38%

d)<em> In order to approximate a binomial distribution with a normal distribution we need a large sample like the one taken in c).</em>

In general, we can do this if the sample of size n the following inequalities hold:

np\geq 5 \;and\;nq \geq 5

in our case np = 200*0.85 = 170 and nq = 200*0.15 = 30

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Answer:

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The number of points in the residual plot is always equal to those in the scatterplot.

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Answer is A.

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