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Verdich [7]
2 years ago
14

I uploaded the answer to a file hosting below

Mathematics
1 answer:
lutik1710 [3]2 years ago
3 0

Answer:

C. is awnser

Step-by-step explanation:

Can I have brainliest * if wrong then D

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What is the difference between a histogram and a cumulative histogram?
diamong [38]

Answer: The answer is provided below

Step-by-step explanation:

A histogram is a diagram which consist of rectangles whereby the area is proportional to frequency of a variable and the width is equal to class interval. A histogram is a commonly used graph that is used to show frequency distributions.

The cumulative histogram is a histogram whereby the vertical axis doesn't gives only the counts for a single bin, but gives the counts for that bin and all the bins for the maller values of a response variable.

Cumulative histograms are similar to normal histograms, but the main difference is that they graph cumulative frequencies unlike histograms that graph just frequencies.

3 0
3 years ago
As part of a screening process, computer chips must be operated in an oven at 145 °C. Ten minutes after starting, the temperatur
Novosadov [1.4K]

Answer:

Step-by-step explanation:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math.  It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

y=Ce^{kt}

Filling in our formula with the 2 conditions we are given:

65=Ce^{10k}   and   85=Ce^{15k}

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k.  If

65=Ce^{10k} then

\frac{65}{e^{10k}}=C which, by exponential rules is the same as

C=65e^{-10k}

Since that value of C is the same as the value of C in the other equation, we sub it in:

85=65e^{-10k}(e^{15k})

Divide both sides by 65 and use the rules of exponents again to get

\frac{85}{65}=e^{-10k+15k} which simplifies down to

\frac{85}{65}=e^{5k}

Take the natural log of both sides to get

ln(\frac{85}{65})=5k

Do the log thing on your calculator to get

.2682639866 = 5k and divide both sides by 5 to find k:

k = .0536527973

Now that we have k, we sub THAT value in to one of the original equations to find C:

65=Ce^{10(.0536527973)}

which simplifies down to

65=Ce^{.536527973}

Raise e to that power on your calculator to get

65 = C(1.710059171) and divide to solve for C:

C = 38.01038064

Now sub in k and C to the final problem when t = 23:

y=38.01038064e^{(.0536527973)(23)} which simplifies a bit to

y=38.01038064e^{1.234014338}

Raise e to that power on your calculator to get

y = 38.01038064(3.434991111) and

finally, the temp at 23 minutes is

130.565

6 0
3 years ago
Through(-3,0)and (0,3)
larisa86 [58]

Answer:

Step-by-step explanation:

the  equation line passes through(-3,0)and (0,3)  is : y=ax+b

a is the slope   a = (3-0)/(0+3) = 1

so : y= x +b  calculate b  line passes through(-3,0)and (0,3)

x=0   y=3  : 3 = 0+b       so  b= 3

this equation is : y= x+3

4 0
3 years ago
Given: F(x)=2x-1; g(x)=3x+2; h(x)=x^2 find h(x+a)-h(x)
umka21 [38]
h(x)=x^2\\\\h(x+a)=(x+a)^2=x^2+2ax+a^2\\\\h(x+a)-h(x)=(x^2+2ax+a^2)-(x^2)=2ax+a^2
5 0
3 years ago
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Can you please my math people
valentina_108 [34]

Answer:

Step-by-step explanation:

5 0
2 years ago
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