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3 years ago

PLEASE HELP

Mathematics

2 answers:

snow_lady [41]3 years ago

8 0

Answer:

1. 17 to 12, 17:12, 17/12

2. 1 to 9, 1:9, 1/9

3. one fifth, two tenths and three over fifteen

4. 3/4 or 3 to 4 or 3:4

Step-by-step explanation:

1. To write a ratio between x and y, in this order, we use the three forms :

x to y

Then, if a zoo has 17 monkeys and 12 chimpanzees the relationships are:

17 to 12, 17:12, 17/12

2. The relationship is between 1 home for every 9 homes. The 4 televisions doesn't matter, that's just the description. Then, the ratios are:

1 to 9, 1:9, 1/9

3. We want three ratios equal to 7/35. Dividing numerator and denominator by 7 we get: 1/5 (one fifth). Multiplying this result by 2, we get: two tenths; and multiplying it by 3, we get: three over fifteen

4. The ratio between Number of tickets Jeanne sold (21) and the number of tickets Jack sold (28) is: 21/28, Dividing numerator and denominator by 7 we get: 3/4

elixir [45]3 years ago

6 0

Answer:

Ratio states that it shows the relative sizes of two or more values.

i,e a to b or a: b or $\frac{a}{b}$

(1)

Given the statement: A zoo has 17 monkeys and 12 chimpanzees

Write this situation in three ways, comparing the first quantity to the second quantity.

By definition of ratio:

Ratio in three ways are:

17 to 12

17 : 12

$\frac{17}{12}$ or we can write this as 17 over 12.

therefore, Option B is correct.

(2)

As per the statement: In Largo City, about one in every nine homes contains four televisions

Write this situation in three ways in the form of ratio, comparing the first quantity to the second quantity.

By definition of ratio:

Ratio in three ways are:

4 to 9

4 : 9

$\frac{4}{9}$

therefore, only options C is correct.

(3)

write three ratios for 7 over 35

By definition of ratio;

We have;

$\frac{7}{35}$

Divide both numerator and denominator by 7 we have;

$\frac{1}{5}$ .....[A]

⇒ $\frac{7}{35} =\frac{1}{5}$

Multiply both numerator and denominator by 2 in A we get;

$\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{1}{10}$

Similarly,

Multiply both numerator and denominator by 3 in A, we get

$\frac{1}{5} = \frac{1 \times 3}{5 \times 3} = \frac{3}{15}$

Therefore, 3 ratios equal to 7 over 35 are;

$\frac{1}{5}$ or we can write it as one-fifth

$\frac{2}{10}$ or two -tenths

$\frac{3}{15}$ or three over fifteen

hence, Option D is correct.

(4)

Given the statement: Jake sold 28 tickets to the school fair, and Jeanne sold 21 tickets

Number of ticket Jake sold = 28 ticket.

Number of ticket Jeanne sold = 21 ticket.

To find: Number of ticket Jeanne sold to number of ticket Jake sold.

By definition of ratio

⇒ $\frac{21}{28}$

Therefore, the ratio in simplest form is, $\frac{21}{28}$

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Find the percent of discount for $70 discounted to $63. Round to the nearest percent.

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70 x P = 63
P=63/70 = 0.9
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1-discount= 0.9
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Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of special interest is t

STatiana [176]

Answer:

Part a) Probability that a moose in that age group is killed by a wolf

P (0.5 year) = 0.361

P (1 - 5) = 0.159

P (6 - 10) = 0.267

P (11 - 15) = 0.203

P (16 - 20) = 0.010

Part b)

Expected age of a moose killed by a wold

μ = 5.61 years

Stantard deviation of the ages

σ = 4.97 years

Explanation:

1) Start by arranging the table to interpret the information:

Age of Moose in years Number killed by woolves

Calf (0.5) 107

1 - 5 47

6 - 10 79

11 - 15 60

16 - 20 3

You can now verify the total number of moose deaths identified as wolf kills: 107 + 47 + 79 + 60 + 3 = 296, such as stated in the first part of the question.

2) First question: a) For each age, group, compute the probability that a moos in that age group is killed by a wolf.

i) Formula:

Probability = number of positive outcomes / total number of events.

ii) Probability that a moose in an age group is killed by a wolf = number of moose killed by a wolf in that age group / total number of moose deaths identified as wolf kills.

iii) P (0.5 year) = 107 / 296 = 0.361

iv) P (1 - 5) = 47 / 296 = 0.159

v) P (6 - 10) = 79 / 296 = 0.267

vi) P (11 - 15) = 60 / 296 = 0.203

vii) P (16 - 20) = 3 / 296 = 0.010

3) Second question: b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages.

i) Class midpoint

0.5 0.5

1 - 5 (1 + 5) / 2 = 3

6 - 10 (6 + 10) / 2 = 8

11 - 15 (11 + 15) / 2 = 13

16 - 20 (16 + 20) = 18

ii) Expected age of a moose killed by a wolf = mean of the distribution = μ

μ = Sum of the products of each probability times its age (mid point)

μ = 0.5 ( 0.361) + 3 ( 0.159) + 8 ( 0.267) + 13 ( 0.203) + 18 ( 0.010) = 5.61 years

μ = 5.61 years ← answer

iii) Stantard deviation of the ages = σ

σ = square root of the variace

variance = s = sum of the products of the squares of the differences between the mean and the class midpoint time the probability.

s = (0.5 - 5.61)² (0.361) + (3 - 5.61)² ( 0.159) + (8 - 5.61)² ( 0.267) + (13 - 5.61)² ( 0.203) + (18 - 5.61)² ( 0.010) = 24.65

σ = √ (24.65) = 4.97 years ← answer

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Nookie1986 [14]

Answer:

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y=(6+0+6)/3=4

centroid=(4,4)

x=(3+5-2)/3=2

y=(3-1+1)/3=1

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Serggg [28]

Answer:

Step-by-step explanation:

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