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Setler [38]
3 years ago
12

Please help I need to pass to graduate

Mathematics
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer:

c

Step-by-step explanation:

i just took the test

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Put the following in order for the most area in the tails of the distribution. ​(a) Standard Normal Distribution ​(b) Student's​
tensa zangetsu [6.8K]

Answer:

(b) (c) (a)

Step-by-step explanation:

Standard Normal distribution has a higher peak in the center, with more area in this región, hence it has less area in its tails.

Student's​ t-Distribution has a shape similar to the Standard Normal Distribution, with the difference that the shape depends on the degree of freedom. When the degree of freedom is smaller the distribution becomes flatter, so it has more area in its tails.

Student's​ t-Distributionwith 1515 degrees of freedom has mores area in the tails than the Student's​ t-Distribution with 2020 degrees of freedom and the latter has more area than Standard Normal Distribution

5 0
3 years ago
Given: (3.2 x 103) + (5.78 x 105) Solve the problem. Show and explain all of your work. ??????
qaws [65]
Simple....

you have: (3.2*103)+(5.78*105)

You want to simplify it..but first remember your order of operations...

1.) Remove the parentheses-->>

(3.2*103)=329.6

and

(5.78*105)=606.9

Which leaves you with...

329.6+606.9=

=936.5

Thus, your answer.
7 0
3 years ago
PLS HELP ME GIVING 43 POINTS I GIVE BRANLIEST PLS HELP ME FOR THIS TEST ITS MATH SUPER EZ
Naya [18.7K]

Answer:lydra=21,horatio=17

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
If |u| = 10, |v| = 8, and the angle formed between them is 60°, then u · v = ?
Gre4nikov [31]
U . v  = |u| . |v| cos 60   = 10 * 8 * 0.5 = 40   This is called the scalar (or dot) product.
5 0
3 years ago
A cold drink is poured out at 52°F. After 2 minutes of sitting in a 72°F room, its temperature has risen to 55°F. Find an equati
I am Lyosha [343]

Answer:

The model for the temperature of the drink can be written as

T=72-20e^{-0.08t}

Step-by-step explanation:

For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

\frac{dT}{dt}=k*(T_r-T)

If we rearrange and integrate

\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}

We know that at time 0, the temperature of the drink was 52°F. Then we have:

T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20

We also know that at t=2, T=55°F

T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08

The model for the temperature of the drink can be written as

T=72-20e^{-0.08t}

7 0
3 years ago
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