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OleMash [197]
3 years ago
7

in slope intercept form write the equation of a line that is parallel to 3x - 5y = 7 and passes through (0, -6)​

Mathematics
1 answer:
lisov135 [29]3 years ago
7 0

Answer:

The gradient is 3/7 make y the subject. Then put the coordinates in the equation.

Step-by-step explanation:

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Calculate GPA of following student. Answer using 4 decimal digits
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Answer:

Step-by-step explanation:

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3 years ago
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PLZ HELP ME <br> I appreciate any help, especially if you can show me how you got the answer
Arte-miy333 [17]

Answer:

C

Step-by-step explanation:

Since this is an indererminate form, use L'Hopital

d(sint)/dt = cos(t)

d[ln(2e^t) - 1] = (2e^t)/[2e^t - 1]

As t --> 0,

cos(0) = 1

(2e^t)/[2e^t - 1] = 2

1/2 is the limit

5 0
3 years ago
Find the mean of the following set of data. Round your answer to two decimal places.
Vera_Pavlovna [14]

Answer:

41.23

Step-by-step explanation:

Mean = (23 + 50 + 49 + 48 + 49 + 32 + 37 + 40 + 41 + 42 + 41 + 41 + 43) ÷ 13

Mean = 536 ÷ 13 = 41.23 (rounded up to two decimal places)

3 0
3 years ago
Read 2 more answers
What is the vertex form of 2x^2+10x-5
Nataly_w [17]
<span><span>Two Solutions
1. x =(10-√140)/4=(5-√<span> 35 </span>)/2= -0.458</span><span> 

2. x =(10+√140)/4=(5+√<span> 35 </span>)/2= 5.458</span></span>
7 0
3 years ago
From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ
bulgar [2K]
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,0)0.1^0(1-0.1)^{3-0}
=1(1)(0.729)
=0.729

(b) x=2
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,2)0.1^2(1-0.1)^{3-2}
=3(0.01)(0.9)
=0.027

(c) x &ge; 2
P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}
=P(2)+P(3)
=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}
=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}
=3(0.01)(0.9)+1(0.001)1
=0.027+0.001
=0.028


8 0
3 years ago
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