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IrinaK [193]
3 years ago
14

I NEED AN EXPLANATION PLS

Mathematics
1 answer:
Serjik [45]3 years ago
8 0
Info:
15 students in Miss Brown’s class
Average score is 95
Most points is 100

What is the lowest score a student could had gotten?

95 divided 5 = 19

So therefore, 19 could be the lowest score
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Which of the following is a solution to the system of two equations y = 3x – 1 and y = 2x + 3?
NeTakaya
Point form: (4,11)
equation form: x=4 y=11
4 0
3 years ago
What is:<br> 9e−7=7e−11<br> Show Work
dexar [7]

Answer: -2

Step-by-step explanation:

9e-7=7e-11

-7e         -7e

2e-7=-11

   +7  +7

2e=-4

/2   /2

e=-2

7 0
3 years ago
What is the GCF of 16 and 72?
MrRissso [65]
The GCF for 16 and 72 is 8, as that is the largest multiple they can both be multiples by.
4 0
3 years ago
I’m so lost i have no idea how to get the function please see pic
Sunny_sXe [5.5K]

You should first recognize the shape of the curve - it's an exponential function, so its equation takes the form y(x)=a\cdot b^{cx} for some constants <em>a</em>, <em>b</em>, and <em>c</em>.

The curve lies above the <em>x</em>-axis, so <em>a</em> must be positive.

The curve is defined everywhere (there are no discontinuities), so <em>b</em> must be positive.

As you move left to right, the function is increasing, so <em>c</em> must also be positive. But to make things simpler, let's assume <em>c</em> = 1.

When <em>x</em> = 0, the curve passes through the point (0, 1). In our equation, we have

y(0) = a\cdot b^0 = a

so it follows that <em>a</em> = 1.

When <em>x</em> = 1, the curve approximately passes through the point (1, 4); so

y(1) = b^1 = 4

so <em>b</em> = 4.

Then the equation of the curve might be \boxed{y = 4^x}.

7 0
3 years ago
The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbul
pshichka [43]

Answer:

P₂   [ 32  ≤ X ≤ 46 ] = 47,71 %

Step-by-step explanation:

The rule:

68-95-99.7

establishes

P₁  [ μ₀ - σ ≤ X ≤ μ₀ + σ] ≈ 0,683

P₂   [ μ₀ - 2σ ≤ X ≤ μ₀ + 2σ] ≈ 0,954

P₃  [ μ₀ - 3σ ≤ X ≤ μ₀ + 3σ] ≈ 0,997

For our paticular case we have

μ₀   =  46

 σ    = 7

Then we get:

μ₀ - σ  = 39            μ₀ +  σ  =   53

μ₀ - 2σ  =  32        μ₀ +  2σ = 60

μ₀ - 3σ  =  25          μ₀ +  3σ  = 67

We were asked by % of replacement request numbering between 32 and 46.

Normal curve is symmetric therefore μ₀ - 2σ  =  32  is just half of the interval  P₂   [ μ₀ - 2σ ≤ X ≤ μ₀ + 2σ] ≈ 0,954, then between 32 and 46 the porcentage of lightbulb is 0,954/2  , is 0,4771 or 47,71 %

P₂   [ μ₀ - 2σ ≤ X ≤ μ₀ ]    = P₂   [ 32  ≤ X ≤ 46 ] = 47,71 %

6 0
4 years ago
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