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muminat
3 years ago
7

What is the measurement of angle 3?

Mathematics
1 answer:
frez [133]3 years ago
7 0

Answer:

Step-by-step explanation:

angle 3 is also (2x-5) degree

if you are wanting the correct number ...i honestly don't know

yet i hope this answer helps

You might be interested in
Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.5
masya89 [10]

Answer:

z= \frac{3.44-2.54}{0.45}=2

P(X>3.44) = P(Z>2) = 0.025

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the grade points avergae of a population, and for this case we know the following properties

Where \mu=2.54 and \sigma=0.45

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

So we can find the z score for the value of X=3.44 in order to see how many deviations above or belowe we are from the mean like this:

z= \frac{3.44-2.54}{0.45}=2

So the value of 3.44 is 2 deviations above from the mean, so then we know that the percentage between two deviations from the mean is 95% and on each tail we need to have (100-95)/2 = 2.5% , because the distribution is symmetrical, so based on this we can conclude that:

P(X>3.44) = P(Z>2) = 0.025

5 0
3 years ago
What is 36.993 to the nearest thousandth
finlep [7]

The next larger thousandth is 36.994 .
The next smaller thousandth is  36.992 .
Neither of those is any nearer to  36.993
than  36.993  already is.

The last '3' at the end of 36.993 is in the thousandths' place.
There is no more piece of another thousandth after it.
So  36.993 is already on a complete thousandth, and
there's no rounding required.

7 0
3 years ago
Multiply the sum of 7/8 and 3/4 by the differences of 5/36 from 11/13
Anni [7]

Answer:

=-\frac{331}{288}=-1.14930\dots

Step-by-step explanation:

(\frac{7}{8} + \frac{3}{4} )\times (\frac{5}{36} - \frac{11}{13} )\\\\\mathrm{Join}\:\frac{7}{8}+\frac{3}{4}:\quad \frac{13}{8}\\\\=\frac{13}{8}\left(\frac{5}{36}-\frac{11}{13}\right)\\\\\mathrm{Join}\:\frac{5}{36}-\frac{11}{13}:\quad -\frac{331}{468}\\\\=\frac{13}{8}\left(-\frac{331}{468}\right)\\\\=-\frac{13}{8}\times\frac{331}{468}\\\\\mathrm{Cross-cancel\:common\:factor:}\:13\\=\frac{1}{8}\times\frac{331}{36}\\\\=-\frac{1\cdot \:331}{8\times\:36}\\\\=-\frac{331}{288}

4 0
3 years ago
What is the interquartile range of the data set 115, 98, 118, 102, 96, 90, 125, 94, 85, 100, 104
kogti [31]

Hello,

interquartile range would be 21

6 0
2 years ago
Read 2 more answers
Consider a rabbit population​ P(t) satisfying the logistic equation StartFraction dP Over dt EndFraction equals aP minus bP squa
maria [59]

Solution:

Given :

$\frac{dP}{dt}= aP-bP^2$         .............(1)

where, B = aP = birth rate

            D = $bP^2$  =  death rate

Now initial population at t = 0, we have

$P_0$ = 220 ,  $B_0$ = 9 ,  $D_0$ = 15

Now equation (1) can be written as :

$ \frac{dP}{dt}=P(a-bP)$

$\frac{dP}{dt}=bP(\frac{a}{b}-P)$    .................(2)

Now this equation is similar to the logistic differential equation which is ,

$\frac{dP}{dt}=kP(M-P)$

where M = limiting population / carrying capacity

This gives us M = a/b

Now we can find the value of a and b at t=0 and substitute for M

$a_0=\frac{B_0}{P_0}$    and     $b_0=\frac{D_0}{P_0^2}$

So, $M=\frac{B_0P_0}{D_0}$

          = $\frac{9 \times 220}{15}$

          = 132

Now from equation (2), we get the constants

k = b = $\frac{D_0}{P_0^2} = \frac{15}{220^2}$

        = $\frac{3}{9680}$

The population P(t) from logistic equation is calculated by :

$P(t)= \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}$

$P(t)= \frac{132 \times 220}{220+(132-220)e^{-\frac{3}{9680} \times132t}}$

$P(t)= \frac{29040}{220-88e^{-\frac{396}{9680} t}}$

As per question, P(t) = 110% of M

$\frac{110}{100} \times 132= \frac{29040}{220-88e^{\frac{-396}{9680} t}}$

$ 220-88e^{\frac{-99}{2420} t}=200$

$ e^{\frac{-99}{2420} t}=\frac{5}{22}$

Now taking natural logs on both the sides we get

t = 36.216

Number of months = 36.216

8 0
3 years ago
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