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3 years ago

Which of the following groups of terms can be used interchangeably when working with normal distributions?

Mathematics

1 answer:

BARSIC [14]3 years ago

5 0

Answer:

The correct answer is D) areas, probability, and relative frequencies

Step-by-step explanation:

This is because each of these terms refers to an amount of space surrounding the mean on the normal distribution curve. Each of them let's us know how likely a spot is to have a value within it.

The z-score would not fit this as it identifies the value and how far away a single spot on the graph would be.

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A grocery store sells a bag of 4 oranges for $2.84. If Arianna spent $3.55 on oranges,

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$4/3=$1.33 per pound.
$9/8 = $1.13 per pound.
The second one is cheaper.

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3 years ago

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The circumference of a circle is 18.84 inches. what is the circle's radius

vivado [14]

Answer:

Circumference : 18.84 inches

Radius: 3 inches

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3 years ago

Pattie is itemizing deductions on her federal income tax return and had $4800

muminat

Answer: 1500

Step-by-step explanation:

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3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago

tres cajas contienen cuatrocientas naranjas la primera tiene veinte naranjas mas q la segunda y treinta mas q la tercera cuantas

Lemur [1.5K]

Answer: Box3 has 120 oranges.

Box1 has 150 oranges.

Box has 130 oranges.

Step-by-step explanation:

I will answer this in English.

The question says:

3 boxes have 400 oranges.

The first one has 20 more than the second and 30 more than the third.

So we have 3 equations:

box1 + box2 + box3 = 400

box1 = box2 + 20

box1 = box3 + 30

Now, we can take the variable box1 in the third equation and replace it on the other two.

box3 + 30 + box2 + box3 = 400

box3 + 30 = box2 + 20

Now we can isolate one of the variables in the second equation, let's isolate box2.

box2 = box3 + 30 - 20 = box3 + 10

now we can replace it in the other equation:

box3 + 30 + box2 + box3 = 400

box3 + 30 + box3 + 10 + box3 = 400

3*box3 + 40 = 400

3*box3 = 400 - 40 = 360

box3 = 360/3 = 120.

Box3 has 120 oranges.

Box1 has 120 + 30 = 150 oranges.

Box has 150 - 20 = 130 oranges.

5 0

3 years ago