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3 years ago

What is 1000 times 10

Mathematics

2 answers:

Setler79 [48]3 years ago

4 0

The answer is 10,000

solmaris [256]3 years ago

3 0

1000 times 10 is 10,000

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Liam is making chocolate chip cookies. The recipe calls for 1 cup of sugar for every 3 cups of flour. Liam only has 2 cups of fl

HACTEHA [7]

Answer:

2/3 cups of sugar

Step-by-step explanation:

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3 years ago

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A class’s test scores are normally distributed if the average score 60 and the standard deviation is 7 choose the spots where th

Brrunno [24]

Answer:

Step-by-step explanation:

We are given that a class's test scores are normally distributed with an average score of 60.

We know that the curve of a normal distribution is symmetric about its mean.

60-14=46

60+14=74

Hence, the point 46 lies to the left of the mean and 74 lies to the right of the mean, and the two points have the same function value.

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3 years ago

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What is the approximate value of X? Round to the nearest tenth.

pychu [463]

Answer

Find out the value of x.

To prove

By using the trignometric identity

$sin\theta = \frac{Perpendicular}{Hypotenuse}$

As given in the diagram.

$\theta = 50 ^{\circ}$

Perpendicular = x

Hypotenuse = 6

Put in the above identity

$sin\ 50^{\circ} = \frac{x}{6}$

$sin\ 50^{\circ} = 0.77\ (Approx)$

Put in the above

$x = 6\times 0.77$

x = 4.6 cm (Approx)

Therefore Option (c) is correct .

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3 years ago

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Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

nadya68 [22]

I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

8 0

3 years ago

A store has apple on sale for $20.00 for 8 pounds. If an apple is approximately 5 ounces, how manyapples can you buy for $120.00

lubasha [3.4K]

Answer: 24 apples

Step-by-step explanation: $20.00 = 8 pounds and 16 ounces = 1 pound So you can do $120.00 divided by 5 = 24 apples

4 0

3 years ago