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max2010maxim [7]
2 years ago
12

-

Mathematics
1 answer:
Whitepunk [10]2 years ago
8 0

Answer:

I hope my answer will helps you.

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95% of students at school weigh between 62 kg and 90 kg. Assuming this data is normally distributed, what are the mean and stand
stiks02 [169]
We can calculate (62 + 90) / 2 = 76. it's <span>known that 95% lies between the mean and twice the standard deviation. </span><span>if the mean is "μ" and the standard deviation is "σ": </span><span>μ-2σ and μ+2σ. i</span><span>f μ=76, then </span><span>76+2σ = 90, so 2σ = 14, so σ = 7.
</span>
<span>the mean is 76 and the standard deviation 7. </span>
5 0
3 years ago
Read 2 more answers
Can anyone help with this?
Morgarella [4.7K]

Answer:

111 degrees

Step-by-step explanation:

the total degrees in a hexagon is 720 degrees.

the known angles add up to 379, so there is 341 degrees left.

angle KFE is 105 degrees because it adds up to 180 degrees with GFK.

angle LAD is 130 degrees because it adds up to 180 degrees with BAD.

130+105=230, so there is now 111 degrees left and that is angle x

5 0
3 years ago
In the 2009 to 2010 school year in country a there were 127,004 in students from country B. This number is 24% more than the num
adoni [48]
If x represents the number of students from country C.

then 122% of x = 116,000

122/100 * x = 116000

Transposing the formula for x;

therefore x = 116000 * 100/122 = 95,082 students
8 0
2 years ago
Plz help me help me help me
exis [7]
The answer is this :) hope I helped

7 0
3 years ago
Read 2 more answers
Let y′′′−9y′′+20y′=0. find all values of r such that y=erx satisfies the differential equation. if there is more than one correc
DerKrebs [107]

we are given

differential equation as

y'''-9y''+20y'=0

we are given

y=e^{rx}

Firstly, we will find y' , y'' and y'''

those are first , second and third derivative

First derivative is

y'=re^{rx}

Second derivative is

y''=r*re^{rx}

y''=r^2e^{rx}

Third derivative is

y'''=r^2*re^{rx}

y'''=r^3e^{rx}

now, we can plug these values into differential equation

and we get

r^3 e^{rx}-9r^2 e^{rx}+20re^{rx}=0

now, we can factor out common terms

e^{rx}(r^3 -9r^2 +20r)=0

we can move that term on right side

(r^3 -9r^2 +20r)=0

now, we can factor out

r(r^2 -9r +20)=0

r(r-5)(r-4)=0

now, we can set them equal

r=0

r-5=0

r=5

r-4=0

r=4

so, we will get

r=0,4,5...............Answer

4 0
3 years ago
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