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s2008m [1.1K]
3 years ago
10

Does anyone know any of these?? please help me!!!

Mathematics
2 answers:
LenKa [72]3 years ago
6 0
2. 435%
3. <span>66.6 %
5. </span><span>17 %
6. .5%
7. 300%
9. </span><span>62.5 %

</span>
marshall27 [118]3 years ago
4 0
5. 0.17 ⇒ 17% 

Hoped this helped.

~Bob Ross ®
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A delivery truck covers a distance of 50 miles every hour. Write the equation that represents, d, the truck has covered over h h
Mnenie [13.5K]

Answer:

50 times h=d

Step-by-step explanation:

4 0
3 years ago
The area under a particular normal curve between 6 and 8 is 0.695. A normally distributed variable has the same mean and standar
TiliK225 [7]

Answer:

69.5%

Step-by-step explanation:

A feature of the normal distribution is that this is completely determined by its mean and standard deviation, therefore, if two normal curves have the same mean and standard deviation we can be sure that they are the same normal curve. Then, the probability of getting a value of the normally distributed variable between 6 and 8 is 0.695. In practice we can say that if we get a large sample of observations of the variable, then, the percentage of all possible observations of the variable that lie between 6 and 8 is 100(0.695)% = 69.5%.

4 0
3 years ago
Maggie has 4 3/5 of icing to decorate cakes for a bake sale. Each container will cover 20 1/4 in2. How many square inches will i
ExtremeBDS [4]

Answer:

93\frac{3}{20} \ square \ inches.

Step-by-step explanation:

Given:

Maggie has 4 3/5 container of icing to decorate cakes for a bake sale.

Each container will cover 20 1/4 square inches.

Question asked:

How many square inches will it cover if she uses all of her icing ?

Solution:

Total number of container Maggie has = 4\frac{3}{5}

Each container will cover = 20\frac{1}{4} square inches.

Now, we can find easily that how many square inches it will cover by using all of her icing by unitary method:

Each container will cover = 20\frac{1}{4}

4\frac{3}{5} container will cover = 20\frac{1}{4}\times4\frac{3}{5} =\frac{20\times4+1}{4} \times\frac{5\times4+3}{5} \

                                                        =\frac{81}{4} \times\frac{23}{5} =\frac{1863}{20} =93\frac{3}{20}\ square\ inches

Thus,  it will cover 93\frac{3}{20} \ square \ inches ,if she uses all of her icing.

7 0
3 years ago
PLEASE HEEEELPPPP ASAP
skelet666 [1.2K]

Answer:

The solution of the first image is: b = √48

The solution of the second image is: c = √125

Step-by-step explanation:

Here we have two triangle rectangles, first, we need to remember the Pythagorean theorem.

For a triangle rectangle with cathetus A and B, and a hypotenuse H, we have the relationship:

A^2 + B^2 = H^2

Where H is the side that is opposite to the right angle (the angle of 90°)

In the first image, we can see that the hypotenuse is equal to 8, and one cathetus is equal to 4.

We want to find the value of b, that is the other cathetus.

Then we have:

4^2 + b^2 = 8^2

b^2 = 8^2 - 4^2

b^2 = 48

b = √48

Second image:

in this case, c is the hypotenuse, a and b are the cathetus.

We know that:

a = 5, b = 10

Then we have the equation:

a^2 + b^2 = c^2

Now we can replace the above values:

5^2 + 10^2 = c^2

25 + 100 = c^2

125 = c^2

√125 = c

5 0
3 years ago
Let X have the uniform distribution U(0, 2) and let the conditional distribution of Y , given that X = x, be U(0, x). Find the j
Elina [12.6K]

Answer:

f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x

E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \

Step-by-step explanation:

We have two random variables X and Y. X \sim Unif(0,2) and given that X=x, Y has uniform distribution (0,x)

From the definition of the uniform distribution we have the densities for each random variable given by:

f_X (x) =\frac{1}{2} , 0\leq x\leq 2

f_{Y|X} (y|x) = \frac{1}{x}, 0\leq y \leq x

And on this case we can find the joint density with the following formula:

f(x,y) = f_{Y|X}(y|x) f_X (x)

And multiplying the densities we got this:

f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x

Now with the joint density we can find the expected value E(Y|x) with the following formula:

E(Y|x) = \int y f_{Y|X}(y|x)dx

And replacing we got:

E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \

5 0
4 years ago
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