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babunello [35]
3 years ago
12

A Park ranger tags 100 animals. Use a proportion to estimate the total population for each sample.

Mathematics
1 answer:
7nadin3 [17]3 years ago
7 0

Answer:

It is given that A Park ranger tags 100 animals.

This means the proportion of animals tagged= \dfrac{100}{x}

where 'x' represents the total population of animals.

We have to find total number of animals in each parts.

1)

23 out of 100 animals are tagged.

That means the proportion of animals tagged= \dfrac{23}{100}

Now we know that the proportion must be equal.

 \dfrac{23}{100}=\dfrac{100}{x}

⇒ x=\dfrac{100\times 100}{23}=434.783

which is approximately equal to 435.

Hence, the total population =435.

3)

8 out of 116 animals are tagged

That means the proportion of animals tagged= \dfrac{8}{116}

Now we know that the proportion must be equal.

 \dfrac{8}{116}=\dfrac{100}{x}

x=\dfrac{100\times 116}{8}\\\\x=\dfrac{11600}{8}=1450

which is approximately equal to 1450

Hence, the total population =1450.

4)

5 out of 63 animals are tagged

That means the proportion of animals tagged= \dfrac{5}{63}

Now we know that the proportion must be equal.

 \dfrac{5}{63}=\dfrac{100}{x}

x=1260

Hence, the total population =1260.

5)

4 out of 83 animals are tagged

That means the proportion of animals tagged= \dfrac{4}{83}

Now we know that the proportion must be equal.

 \dfrac{4}{83}=\dfrac{100}{x}

x=\dfrac{100\times 83}{4}\\\\x=2075

Hence, the total population =2075.

6)

3 out 121 animals are tagged

That means the proportion of animals tagged= \dfrac{3}{121}

Now we know that the proportion must be equal.

 \dfrac{3}{121}=\dfrac{100}{x}

x=\dfrac{100\times 121}{3}\\\\x=4033.33

Hence, the total population =4033.

7)

83 out of 125 animals are tagged

That means the proportion of animals tagged= \dfrac{83}{125}

Now we know that the proportion must be equal.

 \dfrac{83}{125}=\dfrac{100}{x}

x=\dfrac{100\times 125}{83}\\\\x=150.6024

Hence, the total population =151.

8)

7 out 165 animals are tagged

That means the proportion of animals tagged= \dfrac{7}{165}

Now we know that the proportion must be equal.

 \dfrac{7}{165}=\dfrac{100}{x}

x=\dfrac{100\times 165}{7}\\\\x=2357.143

Hence, the total population =2357.

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Answer:

C

Step-by-step explanation:

(a)

2 : 4 ( divide both values by 2)

= 1 : 2

3 : 6 ( divide both values by 3 )

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Thus 2 : 4 = 3 : 6

(b)

3 : 9 ( divide both values by 3 )

= 1 : 3

Thus 1 : 3 = 3 : 9

(c)

4 : 6 ( divide both values by 2 )

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(d)

10 : 2 ( divide both values by 2 )

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Thus 5 : 1 = 10 : 2

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Answer:

The probability of eating pizza given that a new car is bought is 0.952

Step-by-step explanation:

This kind of problem can be solved using Baye’s theorem of conditional probability.

Let A be the event of eating pizza( same as buying pizza)

while B is the event of buying a new car

P(A) = 34% = 0.34

P(B) = 15% = 15/100 = 0.15

P(B|A) = 42% = 0.42

P(B|A) = P(BnA)/P(A)

0.42 = P(BnA)/0.34

P(B n A) = 0.34 * 0.42 = 0.1428

Now, we want to calculate P(A|B)

Mathematically;

P(A|B = P(A n B)/P(B)

Kindly know that P(A n B) = P(B n A) = 0.1428

So P(A|B) = 0.1428/0.15

P(A|B) = 0.952

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3 years ago
Ahmad has $x. Abu has $5 more than Ahmad. Razak has twice as much as Abu. Together they have $175. What is the value of x?
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Abu X + 5

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Together 4x + 15

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(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop
zalisa [80]

Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})

Compute the sample sizes as follows:

MOE=z_{\alpha/2}\times\sqrt{\frac{2\times\hat p(1-\hat p)}{n}

       n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}

          =\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

\hat p_{1}=0.45\\\hat p_{2}=0.58

Compute the sample size as follows:

MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}

       n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}

          =\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666

Thus, the sample sizes are 6666.

7 0
3 years ago
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