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2 years ago

Ava said "There are only three hundred 3- diget numbers" is her statement true or faulse

Mathematics

1 answer:

Eduardwww [97]2 years ago

3 0

Answer: False, there are actually 900 different three-digit numbers

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Explanation:

The three digit numbers span from 100 to 999, including both endpoints.

This means we have 999-100+1 = 900 different three-digit numbers.

You subtract the endpoints (large-small) and add 1 to include the lower endpoint.

Here's a smaller example of why this works: say you had the set {1,2,3,4} and we wanted to count the number of items in this set. Clearly there are 4 items. Note how subtracting the endpoints 4-1 gets us 3 instead, so we add on 1 to include that left endpoint.

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Reptile [31]

Answer:

The figures are congruent because a 270° rotation about the origin and then a reflection over the x-axis will map ΔABC onto ΔLMN.

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2 years ago

Read 2 more answers

The quotient of 15 and a number is 1 over 3 written as an equation

Agata [3.3K]

Answer:

15/x=1/3

Step-by-step explanation:

The quotient of 15 and a number=15/x=1/3

15/x=1/3

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2 years ago

The batting Wang Xiu Ying uses to fill quilts has a thermal conductivity rate of 0.030.030, point, 03 watts (\text{W})(W)left pa

alexgriva [62]

Answer:

0.0003W/cm°C

Step-by-step explanation:

The question is not properly written. Here is the correct question.

The batting wang xiu ying uses to fill quilts has a thermal conductivity rate of 0.03 watts (W) per meter(m) per degree celsius. what is the batting thermal conductivity when w/cm•c

Given the thermal conductivity in W/m°C to be 0.03W/m°C

We are to rewrite the value in W/cm°C

The difference is the unit. The only thing we need to do is to simply convert the unit (metres) in W/m°C to centimeters (cm)

Since 100cm = 1m, 0.03W/m°C can be expressed as shown below;

= 0.03W/m°C

= 0.03 × W/1m×°C

Note that 1m = 100cm, substituting this conversion into the expression, it will become;

= 0.03 × W/100cm × °C

= 0.03/100 × W/cm°C

= 0.0003W/cm°C

Hence the battling thermal conductivity in W/cm°C is 0.0003W/cm°C

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2 years ago

Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu

Svetllana [295]

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of $P(Z$ , you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

4 0

2 years ago

To increase sales, a local donut shop began putting an extra donut in some of the boxes. Customers are unaware of which boxes ha

Bingel [31]

Answer:

Step-by-step explanation:

Hello!

Part A

First, determine your study variable:

X: Number of boxes with an extra donut in a sample of eight.

To see if the variable has a Binomial distribution you have to check if the binomial criteria are met:

1. The number of observation of the trial is fixed (In this case n = 8, the boxes each customer bought make the sample)

2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial (In this case, the amount of donuts in one box does not affect on the probability of the next box having an extra donut)

3. The probability of success in the same from one trial to another (In this case our "success" will be that the box has an extra donut, according to the owners claim that is 1/7; ρ=0,14)

So X≈ Bi (n;ρ)

Where n represents the sample (n=8) and ρ is the probability of success (ρ=0.14)

Part B

The mean of the binomial distribution is E(X)= nρ

E(X)= 8 * 0.14= 1.12

The mean of the distribution is also called the expected value. You'd expect that 1.12 boxes have an extra donut.

The variance of the binomial distribution is V(X)= nρ(1 - ρ)

V(X)= 8*0.14*(1 - 0.14)= 0.9632

Its square root is the standard deviation

√V(X)= 0.98

The standard deviation is a measure of dispersion, it indicates how much the values of the distribution of the central value are separated. In this case, 0.98 indicates that the distribution of the number of boxes with an extra donut is far from the expected value.

Part C

Two of the eight customers buy a box with an extra donut, symbolically:

P(X=2) = P(X≤2) - P(X≤1)= 0.91 - 0.68 = 0.22

There is a 22% chance that two customers bought a box with an extra donut.

Compute:

P(X≥2)= 1 - P(X<2)= 1 - P(X≤1)= 1 - 0.68= 0.32

There is a 32% chance that two or more customers bought a box with an extra donnut.

I hope it helps!

3 0

2 years ago