A circle fits inside a semicirlce with a diameter of 24mm calculate the shaded area

The highest recorded temperature during the month of July for a given year in Death Valley, California has an approximately norm

S_A_V [24]

Answer:

$P(X$

And we can find this probability using excel or the normal standard tabe and we got:

$P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the temperatures of a population, and for this case we know the distribution for X is given by:

$X \sim N(123.8,3.1)$

Where $\mu=123.8$ and $\sigma=3.1$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard tabe and we got:

$P(z$

6 0

2 years ago

The population of Rock Springs is growing at a constant rate. The expression represents the increase in population in x years.

irakobra [83]

The function is given by

y=45,200 + 1,900x

and is of the form y = mx+b , where m is the rate of change and b is the fixed value ( or we can say initial value)

Now when we compare we get m = 1,900 for our problem .

So the population increases every year by 1,900.

3 0

2 years ago

Read 2 more answers

There were 340,000 cattle placed on feed. Write an equivalent ratio that could be used to find how many of these cattle were bet

lutik1710 [3]

There were 340,000 cattle placed on feed

How many of the 340,000 cattle placed on feed were between 700 and 799 pounds?

Given the fraction of total cattle for 700 - 799 pounds is 2/5

Let x be the number of cattle between 700 - 799 pounds

We make a proportion using the fraction

$\frac{2}{5} = \frac{x}{340,000}$

Cross multiply it and solve for x

340000* 2 = 5x

680000 = 5x

Divide by 5 on both sides

So x= 136,000

There were 136,000 cattle between 700 and 799 pounds

3 0

1 year ago

Read 2 more answers

In a local ice sculpture contest, one group sculpted a block into a rectangular based pyramid. The dimensions of the base were 3

m_a_m_a [10]

Answer:

1. The amount of ice needed = 18 m²

2. The amount of fabric needed to manufacture the umbrella is 0.76 m²

3. The height of the cone, is 3.75 cm

4. The dimensions of the deck are;

Width = 28/3 m, breadth = 28/3 m

The area be 87.11 m²

5. The dimensions of the optimal design for setting the storage area at the corner, we have;

Width = 10m

Breadth = 10 m

The dimensions of the optimal design for setting the storage area at the back of their building are;

Width = 7·√2 m

Breadth = 7·√2 m

Step-by-step explanation:

1. The amount of ice needed is given by the volume, V, of the pyramid given by V = 1/3 × Base area × Height

The base area = Base width × Base breadth = 3 × 5 = 15 m²

The pyramid height = 3.6 m

The volume of the pyramid = 1/3*15*3.6 = 18 m²

The amount of ice needed = 18 m²

2. The surface area of the umbrella = The surface area of a cone (without the base)

The surface area of a cone (without the base) = π×r×l

Where:

r = The radius of the cone = 0.4 m

l = The slant height = √(h² + r²)

h = The height of the cone = 0.45 m

l = √(0.45² + 0.4²) = 0.6021 m

The surface area = π×0.4×0.6021 = 0.76 m²

The surface area of a cone (without the base) = 0.76 m²

The surface area of the umbrella = 0.76 m²

The amount of fabric needed to manufacture the umbrella = The surface area of the umbrella = 0.76 m²

3. The volume, V, of the cone = 1/3×Base area, A, ×Height, h

The volume of the cone V = 150 cm³

The base area of the cone A = 120 cm²

Therefore we have;

V = 1/3×A×h

The height of the cone, h = 3×V/A = 3*150/120 = 3.75 cm

4. Given that the deck will have railings on three sides, we have;

Maximum dimension = The dimension of a square as it is the product of two equal maximum obtainable numbers

Therefore, since the deck will have only three sides, we have that the length of each side are equal and the fourth side can accommodate any dimension of the other sides giving the maximum dimension of each side as 28/3

The dimensions of the deck are width = 28/3 m, breadth = 28/3 m

The area will then be 28/3×28/3 = 784/9 = $87\frac{1}{9}$ =87.11 m²

5. The optimal design for setting the storage area at the corner of their property with four sides is having the dimensions to be that of of a square with equal sides of 10 m each as follows;

Width = 10m

Breadth = 10 m

The optimal design to have the storage area at the back of their building having a fence on only three sides, is given as follows;

Storage area specified = 98 m²

For optimal use of fencing, we have optimal side size of fencing = s = Side length of a square

s² = 98 m²

Therefore, s = √98 = 7·√2 m

Which gives the width = 7·√2 m and the breadth = 7·√2 m.

8 0

1 year ago

Let P2 be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by 10x2+4xâ1, 3xâ4x2+3, and

lord [1]

I suppose

$H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}$

The vectors that span $H$ form a basis for $P_2$ if they are (1) linearly independent and (2) any vector in $P_2$ can be expressed as a linear combination of those vectors (i.e. they span $P_2$ ).

Independence:

Compute the Wronskian determinant:

$\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0$

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of $H$ is 3.

Span:

Write an arbitrary vector in $P_2$ as $ax^2+bx+c$ . Then the given vectors span $P_2$ if there is always a choice of scalars $k_1,k_2,k_3$ such that

$k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c$

which is equivalent to the system

$\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}$

so the vectors do span $P_2$ .

The vectors comprising $H$ form a basis for it because they are linearly independent.

4 0

2 years ago