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Ostrovityanka [42]

2 years ago

12

You really want to earn an a in a class, but you still need to take one more test. suppose that your previous test scores are 85

, 95, and 79. assume that the test is out of 100 points. what score would you need to get on the last test? explain.

1 answer:

Alex787 [66]2 years ago

7 0

If u get 100 your average will be 89.75 which rounds up to a 90

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A car accelerates at 3.00 m/s2. The car has a mass of 1354kg. What net force will produce the acceleration?

otez555 [7]

Answer:

4062N

Step-by-step explanation:

Using Newton’s Second Law;

F = ma

F = (1354)(3)

F = 4062N

Hope this helps! :D

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1 year ago

4) Look at the underlined digits in the next number. Use the drop-down menus to compare the digit values. 24,480 4) Click on the

Sergio039 [100]

Answer:

Step-by-step explanation:

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3 years ago

Prove that sin²∅+cos²∅=1 ?

sasho [114]

Step-by-step explanation:

Let ABC be a right angled triangle where there's a right angle in B . Let the measure of the angle BAC be ∅. So,

$\sin(∅) = \frac{BC}{AC}$

$= > {\sin}^{2}∅ = \frac{ {BC}^{2} }{ {AC}^{2} }$

Also,

$\cos(∅) = \frac{AB}{AC}$

$= > { \cos}^{2}∅ = \frac{ {AB}^{2} }{ {AC}^{2} }$

Now adding the values of sin^2∅& cos^2∅,

${ \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {BC}^{2} }{ {AC}^{2} } + \frac{ {AB}^{2} }{ {AC}^{2} }$

$= > { \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {AB}^{2} + {BC}^{2} }{ {AC}^{2} }$

But we know that

${AC}^{2} = {AB}^{2} + {BC}^{2}$ by applying Pythagorean Theorem

So ,

$= > { \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {AC}^{2} }{ {AC}^{2} } = 1$

7 0

2 years ago

If (3 , K) and (2, 4k) are two points on a graph of a line and k is not equal to 0, what is the slope of the line?

agasfer [191]

I honestly don’t know, Try to do Y2-Y1 and X2-X1 to get your answer.

7 0

2 years ago

there has to be at least one operating path. Once a component fails on a path, that path is no longer operating. The reliability

Eva8 [605]

Answer:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$

Step-by-step explanation:

We can assume that the system work in series

If we have in general n units the reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. fror n units must succeed for the system to succeed. The reliability of the system is then given by:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$

8 0

2 years ago