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3 years ago

CAN somebody plz plz help me with raios

Mathematics

1 answer:

Kaylis [27]3 years ago

5 0

I can help what are the problems

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At a store, customers are randomly selected to participate in a survey. On Friday, there were 500 customers at the store. Out of

Svetlanka [38]

Divide the number of people selected by the total number of people.

90/500 = .18

This means that 18% of the customers were selected for the survey.

If the probability is the same on Saturday, then we can multiply the expected customers by our .18

700 x .18 = 126

126 should be selected for the survey on Saturday.

7 0

3 years ago

Order these numbers in order 0.5, 1 1/4, 0.25, 3/8

vodomira [7]

Answer:

0.5,0.25,3/8,1 1/4

Step-by-step explanation:

3 0

3 years ago

Please help asap

FromTheMoon [43]

Answer: $9

Step-by-step explanation:

10 percent of 9 or 90 broken into 10 parts is 9 also 90 divided by 10 is 9

3 0

3 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

4 0

3 years ago

In a basketball game, a regular basket was worth 2 points and a long-distance basket was worth 3 points. if there were 45 basket

IgorLugansk [536]

Let
x: number of regular basketball
y: number of long-distance basket
We have the following system of equations:
2x + 3y = 96
x + y = 45
Solving the system we have
y = 45-x
2x + 3 (45-x) = 96
2x +135 -3x = 96
-x = 96 -135
x = 39
Then,
y = 45-x
y = 45-39
y = 6
answer
were made
regular baskets = 39
long-distance baskets = 6

3 0

3 years ago