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2 years ago

73, 66, 80, 81, 45

Mathematics

1 answer:

Simora [160]2 years ago

5 0

Answer:

Mean-69

Median-73

Mode- There is no mode

Range-36

Step-by-step explanation:

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The domain of the function <br> g(x) = √(x2−3x−4) is? Use the fishball method

frosja888 [35]

Result:
{x<= -1 or x>= 4 and x belongs to real numbers}

6 0

3 years ago

Please I need help. you will get points

iragen [17]

Answer:

t(32) = 94

Step-by-step explanation:

a. I think s is the independent variable

b I think t or t(s) is the dependent variable since the total amount of money depends on how many treats are sold.

t(32) = 30 + 2(32)

t(32) = 30 + 64

t(32) = 94

I'm a little unsure on the dependent and independent variable but I know the function is correct.

Hope this helps :)

7 0

3 years ago

Whats 5 + 5, please I need to know

vlabodo [156]

Answer:

Step-by-step explanation:

Use your fingers

3 0

3 years ago

Read 2 more answers

PLS HELP 15 POINTS. Divide 400 by the frequencies for numbers 1-6. This will give you the relative frequency for each number.

skelet666 [1.2K]

Answer:

i dont know what you mean by this but 400 divided by the numbers 1,2,3,4,5,6 ,okay

Step-by-step explanation:

400÷1=400

400÷2=200

400÷3=133.333333333

400÷4=100

400÷5=80

400÷6=66.6666666667

You're welcome

4 0

3 years ago

ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl

melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

$\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x} \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm$

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

$\frac{AQ}{QP}=\frac{AB}{CP} \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19$

$\frac{BQ}{CQ}=\frac{AB}{CP} \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8$

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

7 0

3 years ago