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777dan777 [17]
3 years ago
11

Y= -1/2 + 5

Mathematics
2 answers:
Andreas93 [3]3 years ago
4 0
W . W is the answer yww <33
DerKrebs [107]3 years ago
3 0

Answer:

w

Step-by-step explanation:

w is the answer

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Please send help, I need it answered.
guapka [62]

Answer:

100 degrees

Step-by-step explanation:

5 0
4 years ago
5x=2x-15<br> y=2x-15<br> and y=5x
Ipatiy [6.2K]
5x = 2x-15
3x = -15
x = -5

substitute x = -5 into one of your equations
so y = 5x becomes y = 5 x -5
y= -25

x=-5 and y=-25
7 0
3 years ago
Read 2 more answers
A master electrician earns $62 per hour. His apprentice earns $40 per hour. The master electrician works 3 hours more than the a
photoshop1234 [79]
X=hours master worked
y=hours apprentice worked


62x+40y=492

if master worked 3hours more than apprntice
x=3+y

sub 3+y for x

62(3+y)+40y=492
expand
186+62y+49y=492
186+102y=492
minus 186 both sides
102y=306
divide both sides by 102
y=3

sub back
x=3+y
x=3+3
x=6

master electrician worked 6 hours
6*62=372

master electrician earned $372
7 0
3 years ago
Read 2 more answers
Whats the answer to this?
mina [271]

instead of P+2x = y it should be P-2x=y

 so then it is P-2x/2 =y

6 0
4 years ago
A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly select
Stels [109]

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; \mu = 57,800  and  \sigma = 750.

Let X = randomly selected element of the population

The z probability is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)  

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( \frac{X-\mu}{\sigma} <= \frac{58000-57800}{750} ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( \frac{X-\mu}{\sigma} < \frac{57000-57800}{750} ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = \frac{\sigma}{\sqrt{n} } = \frac{750}{\sqrt{100} } = 75

The z probability is given by;

           Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)  

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{58000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{57000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

7 0
3 years ago
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