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3 years ago

In a standard normal distribution, what percent of values are less than -0.23?

Mathematics

1 answer:

Vika [28.1K]3 years ago

5 0

Answer:

0.4

Step-by-step explanation:

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Help! :(<br><br> It’s in the pic :3

Harlamova29_29 [7]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

How do I solve and find x for this question

Fed [463]

Answer: -9.1

Explanation
-3x +12.4=39.7
First minus 12.4 from both sides

-3x +12.4=39.7
-12.4 -12.4
-3x=27.3
Now devid both sides by -3
X= -9.1

7 0

2 years ago

'Five times a number minus eight is 17.5.

11Alexandr11 [23.1K]

Answer:

5.1

Step-by-step explanation:

you need to work backwards.

so do 17.5 + 8 then divide by 5 to get 5.1

i hope this helps :)

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2 years ago

The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th

vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by $P(t)=1.394(1.006)^t$

To find the derivative you need to:

$\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t$

To find the population growing at the start of 2014 we say t = 0

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year$

To find the population growing at the start of 2015 we say t = 1

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year$

To convert billion to million you multiple by 1000

$P(0)' = 0.00833901 \:Billion/year \cdot 1000 = 8.34 \:Million/year \\P(1)' = 0.00838904 \:Billion/year \cdot 1000 = 8.39 \:Million/year$

6 0

3 years ago

What is the radius of a circumference 56.5

Bas_tet [7]

8.992254284692088 is the radius I believe

7 0

3 years ago

In a standard normal distribution, what percent of values are less than -0.23?​

In a standard normal distribution, what percent of values are less than -0.23?