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Korolek [52]
3 years ago
10

80 % of _ games is 32 games

Mathematics
2 answers:
ddd [48]3 years ago
8 0
40 games is the answer you're looking for
andrew-mc [135]3 years ago
7 0
80% × x = 32
.80x = 32
x = 32/.8
x = 40
You might be interested in
According to Gallup’s annual poll on personal finances, while most U.S. workers reported living comfortably now, many expected a
tester [92]

Answer:

- The required percentage to 4 d.p. = 59.5126%

- The required percentage in decimal point to 4 d.p. = 0.5951

Step-by-step explanation:

Proportion that said they have enough money to live comfortably now and expected to do so in the future = 55% = 0.55

Sample size = n = 200

The Central Limit Theorem ensures that we can say that:

1) The sample proportion is equal to the population proportion.

p = μₓ = μ = 0.55

2) Standard Deviation of the sampling distribution = σₓ = √[p(1-p)/n] = √(0.55×0.45/200) = 0.0352

3) We can say that the sample distribution approximates a normal distribution especially when

np > 5 and nq > 5 (which is true for this sample size)

The probability is 90% that less than what sample percentage will say they expect to live comfortably

To find this sample percentage, let that that sample percentage be x' and its z-score be z'

z' = (x' - μ)/σₓ

But we are told that

P(z < z') = 90% = 0.90

Using the normal distribution table

z' = 1.282

1.282 = (x' - 0.55)/0.0352

x' = 0.55 + (0.0352×1.282) = 0.5951264

Hence, the required sample percentage = 59.5126% to 4 d.p.

Hope this Helps!!!

7 0
4 years ago
Is this correct not that sure
Arte-miy333 [17]
Yes there is a 2/3 chance. This is fun stuff!
6 0
3 years ago
Solve the inequality 2n – 1 &lt; 39​
MatroZZZ [7]

Answer:

n < 20​

Step-by-step explanation:

2n – 1 < 39​

Add 1 to each side

2n – 1+1 < 39+1

2n < 40

Divide each side by 2

2n/2 < 40/2

n < 20​

3 0
3 years ago
Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
3 years ago
A shop has 15% sale. Originally a computer cost £275. How much does the computer cost in the sale?
lutik1710 [3]
15%×£275=£41.25
£275-£41.25=£233.75
5 0
3 years ago
Read 2 more answers
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