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3 years ago

Plz help this is urgent

Mathematics

1 answer:

Alex777 [14]3 years ago

6 0

Answer:

A. 13 + t = 17

Step-by-step explanation:

The top rectangle shows 13 and t, the 13 and t rectangle is the same size as the 17 rectangle so 13 + t = 17. It's not multiplication as there's no multiple's of 13 or t on the diagram.

You might be interested in

Perform the indicated operation. Be sure the answer is reduced.

Morgarella [4.7K]

The answer is: $k^{2} /2
$
The explanation is shown below:
1. You have the following expression given in the exercise above:
$( 4k^{2}) /(2k+3)(2 k^{2} -5k-12)/(8k-32)
$
2. By applying the distributive property, you obtain:
$(8k^{4}-20k^{3}-48k)/(16k^{2} -40k-96)$
3. Now, you must factor $k^{2}$ in the numerator and $2
$ in the denominator:
$k^{2}(8 k^{2} -20k-48)/2(8 k^{2} -20k-48)$
4. Simplify:
$k^{2} /2$

8 0

3 years ago

Mr. Akika has 30 24-cent and 33-cent stamps all told. The stamps are worth $8.91. How many of each kind of stamp does he have?

riadik2000 [5.3K]

Answer:

24-cent: 11
33-cent: 19

Step-by-step explanation:

I find it easiest to solve these using the variable to represent the number of the most-expensive item. Let x represent the number of 33-cent stamps. Then the total value (in cents) is ...

33x +24(30 -x) = 891

9x = 171 . . . . . . . . . . . . . subtract 24(30), collect terms

x = 19 . . . . . . . . . . . . divide by 9

30-x = 11 . . . . . . . find the number of 24-cent stamps

Mr. Akika had 11 24-cent stamps and 19 33-cent stamps.

_____

<em>Additional comment</em>

You seem to have several of these to solve. They all have a similar solution.

If we let x represent the number of higher-value items, N, the total number of items, V the total value, and v1 and v2 the individual values (v2 > v1), the equation is ...

v2(x) +v1(N -x) = V

x(v2 -v1) = V -v1·N

x = (V -v1·N)/(v2 -v1) . . . . generic solution

Using this formula in the problem here, we find ...

x = (891 -24(30))/(33 -24) = 171/9 = 19 . . . as above

7 0

3 years ago

The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 4.5 per week. Find the probabili

Mariana [72]

Answer:

a) 0.01111

b) 0.4679

c) 0.33747

Step-by-step explanation:

We are given the following in the question:

The number of accidents per week can be treated as a Poisson distribution.

Mean number of accidents per week = 4.5

$\lambda= 4.5$

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

a) No accidents occur in one week.

$P(x =0)\\\\= \displaystyle\frac{4.5^0 e^{-4.5}}{0!}= 0.01111$

b) 5 or more accidents occur in a week.

$P( x \geq 5) = 1-\displaystyle \sum P(x$

c) One accident occurs today.

The mean number of accidents per day is given by

$\lambda = \dfrac{4.5}{7} = 0.64$

$P(x =1)\\\\= \displaystyle\frac{0.64^1 e^{-0.64}}{1!}= 0.33747$

5 0

3 years ago

The area of a rectangular ceiling tile is 756 square inches. The perimeter is 114 inches. What are the dimensions of the tile?

kogti [31]

Here we need to use what we know about rectangles to make a system of equations.

By solving that system we found that the tile has a length of 36 inches and a width of 21 inches.

Remember that for a rectangle of length L and width W, the perimeter is:

P = 2*(L + W)

And the area is:

A = W*L

Here we know that the perimeter is 114 inches, then we can write:

114in = 2*(L + W)

We also know that the area is 756 in^2, then we can write:

756 in^2 = L*W

So we found two equations, which means that we have a system of two equations with two variables:

114in = 2*(L + W)

756 in^2 = L*W

To solve this, the first step is to isolate one of the variables in one of the equations, we can isolate L in the first equation:

114in = 2*(L + W)

114in/2 = (L + W)

57in = L + W

57in - W = L

Now that we have an expression equivalent to L, we can replace it in the other equation to get:

756 in^2 = L*W

756 in^2 = (57in - W)*W

Now we can solve this for W.

756 in^2 = W*57in - W^2

W^2 - W*57in + 756 in^2 = 0

The solutions are given by the Bhaskara's formula:

$W = \frac{57in \pm \sqrt{(-57in)^2 - 4*1*(756in^2)} }{2*1} = \frac{57in \pm 15in}{2}$

Then the two possible values of the width will be:

W = (57in + 15in)/2 = 36 in

W = (57in - 15in)/2 = 21 in

Suppose that we choose the second solution, W = 21in

Now using the equation 57in - W = L we can find the value of L

L = 57in - W = 57in - 21in = 36in

L = 36in

Then we found that the tile has a length of 36 inches and a width of 21 inches.

If you want to learn more, you can read:

brainly.com/question/11137975

4 0

3 years ago

Lelana bought 21 packs of washable crayons and x packs of non-washable crayons. Write an ex- pression to show how many crayons L

scoundrel [369]

Answer:

$21 + x$ packs

Step-by-step explanation:

Given

$Washable \to 21$

$Non \to x$

Required

The total packs bought

This is calculated as:

$Total = Washable + Non$

$Total = 21 +x$

3 0

3 years ago