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Gemiola [76]
3 years ago
10

4200 fans increased by 0.5%

Mathematics
1 answer:
Ronch [10]3 years ago
6 0
I'm only in 4th grade and all these qusition are hard I'm so so so sorry
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Qual o valor que Z assume?<br> -Cz+ 6z = Tz+ 83
o-na [289]

Answer:

\large \boxed{z = \dfrac{83}{6 - C - T}}

Step-by-step explanation:

\begin{array}{rccl}-Cz + 6z & = & Tz + 83 & \\6z -Cz - Tz &= & 83 & \text{Subtracted Tz from each side}\\z(6 - C - T)  & = & 83 & \text{Removed the common factor}\\z& = & \mathbf{\dfrac{83}{6 - C - T}} &\text{Divided each side by 6 - C - T}\\\end{array}\\\\\large \boxed{\mathbf{z = \dfrac{83}{6 - C - T}}}

3 0
3 years ago
Determine the amplitude, period, and phase shift of the function f(x)=cos (8(x-1))+2​
SCORPION-xisa [38]

We have been given a function f(x)=\text{cos}(8(x-1))+2. We are asked to find the amplitude, period and phase shift of the function

We can see that our given function is in form f(x)=A\cdot \cos(B(x-C))+D, where,

|A| = Amplitude.

Period = \frac{2\pi}{|B|}

C = Horizontal shift,

D = Vertical shift.

We can see that value of a is 1, therefore, the amplitude of given function would be 1.

We can see that B is equal to 8, so we will get:

\text{Period}=\frac{2\pi}{|B|}= \frac{2\pi}{|8|}=\frac{\pi}{4}

Therefore, the period of given function is \frac{\pi}{4}.

Since the value of C is 1, therefore, horizontal shift is 1.

Since the value of D is 2, therefore, vertical shift would be 2.

4 0
2 years ago
Consider the distribution of exam scores​ (graded from 0 to​ 100) for 86 students when 38 students got an​ A, 28 students got a​
Bond [772]

<u>Answer:</u>

There is only 1 peak in this distribution.

<u>Step-by-step explanation:</u>

In the distribution of exam scores (graded from 0 100)  for 86 students 38 students got an A, 28 students got a B, and 20 students got a C.

a) Since, 38 > 28 > 20 , there is only 1 peak for this distribution.  

5 0
3 years ago
P(0,-1) that is parallel to the line y= -2x+3​
Mandarinka [93]

Answer:

y=2x+3

A parallel line is only supposed to have an oposite slope.

3 0
3 years ago
Sketch the areas under the standard normal curve over
SOVA2 [1]

Answer:

a) P(Z

b) P(Z>0.15)=1-0.560=0.440

c) P(-1.40

Step-by-step explanation:

a)To the left of z = 0.72.

For this case the shaded area is on the figure 1 attached.

And we can find the area with the normal standard table of with the following excel code:

=NORM.DIST(0.72,0,1,TRUE)

P(Z

b)To the right of z = 0.15.

For this case the shaded area is on the figure 2 attached.

And we can find the area with the normal standard table of with the following excel code:

=1-NORM.DIST(0.15,0,1,TRUE)

P(Z>0.15)=1-0.560=0.440

c)Between z = -1.40 and z = 2.03.

For this case the shaded area is on the figure 3 attached.

And we can find the area with the normal standard table of with the following excel code:

=NORM.DIST(2.03,0,1,TRUE)-NORM.DIST(-1.40,0,1,TRUE)

P(-1.40

5 0
3 years ago
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