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marta [7]
3 years ago
8

Slopes of parallel and perpendicular lines

Mathematics
1 answer:
timurjin [86]3 years ago
8 0

Answer:

y=  x+  3

Step-by-step explanation:

You might be interested in
A biased coin is flipped twice.
Anni [7]

Answer:

0.48.

Step-by-step explanation:

There are 2 ways that a tail lands once:-

Tail + head or

head + tail.

Prob(Tail on one throw) = 0.4

Prob(Head on one throw) = 1 - 0.4 = 0.6.

Prob(Tail first then head) = 0.4 * 0.6 = 0.24

Head then tail has the same probability (0.6*0.4).

So the requred probability = 0.24 + 0.24

= 0.48.

8 0
2 years ago
Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean len
Marina86 [1]

Answer:

Step-by-step explanation:

Hello!

You need to construct a 95% CI for the population mean of the length of engineering conferences.

The variable has a normal distribution.

The information given is:

n= 84

x[bar]= 3.94

δ= 1.28

The formula for the Confidence interval is:

x[bar]±Z_{1-\alpha/2}*(δ/n)

Lower bound(Lb): 3.698

Upper bound(Ub): 4.182

Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242

I hope it helps!

6 0
3 years ago
Answer this question please
astraxan [27]

Answer:

6

Step-by-step explanation:

add 2 each time

3 0
3 years ago
Read 2 more answers
1 point
swat32

Answer:

\log_{10}(147) = 2.1673

Step-by-step explanation:

Given

\log_{10} 3 = 0.4771

\log_{10} 5 = 0.6990

\log_{10} 7= 0.8451

\log_{10} 11 = 1.0414

Required

Evaluate \log_{10}(147)

Expand

\log_{10}(147) = \log_{10}(49 * 3)

Further expand

\log_{10}(147) = \log_{10}(7 * 7 * 3)

Apply product rule of logarithm

\log_{10}(147) = \log_{10}(7) + \log_{10}(7) + \log_{10}(3)

Substitute values for log(7) and log(3)

\log_{10}(147) = 0.8451 + 0.8451 + 0.4771

\log_{10}(147) = 2.1673

3 0
2 years ago
One number is 1 more than 2 times another. Their product is 10. Find the numbers.
Taya2010 [7]
Whenever a problem says "one number" and "another number" we can substitute x and y. for this we have x=2y+1 and xy=1. since we know the value of x (2y+1) we can substitute it for the other equation to get (2y+1)y=10. simplify to get 2y^2+y=10. from here you can do a few methods to solve this, but the simplest in my opinion is by factoring. 

In order to factor it must be equal to 0, so we have 2y^2+y-10=0. We factor this and get y=5 (we also get y=-4 but it is an extraneous root). now we can plug that into either equation and find that x=2.

this means our two numbers are 5 and 2
3 0
3 years ago
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