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3 years ago

Match each equation to its factorized version and solution.

Mathematics

1 answer:

bagirrra123 [75]3 years ago

5 0

Answer:

$24x -{6x}^{2} = 0 \: matches \: with \: 6x(4 - x) \: and \: the \: solution \: x = 0 \ \: or \: \: x = 4$

$2 {x}^{2} + 6x = 0 \: matches \: with \: 2x(x + 3) = 0 \: and \: the \: solution \: x = 0 \: or \: x = - 3$

$4x - {x}^{2} = 0 \: matches \: with \: x(4 - x) = 0 \: and \: the \: solution \: x = 0 \: or \: x = 4$

$14x - {7x}^{2} = 0 \: matches \: with \: 7x(2 - x) = 0 \: and \: the \: solution \: x = 0 \: or \: x = 2$

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Antibiotics in Infancy

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Answer:

e)The information from the sample does not give enough information to support that more than 70% of Canadian children receive antibiotics during the first year of life

Step-by-step explanation:

The proportion we will use for the test is p = 70 %

From a random sample we got

n = 616

x = 438 then p₁ = 438/616 p₁ = 0,71 p₁ = 71 %

Then q₁ = 1 - p₁ q₁ = 1 - 0,71 q₁ = 0,29 q₁ = 29 %

a) Hypothesis test:

Null hypothesis H₀ p₁ = p

Alternative hypothesis Hₐ p₁ > p >70 %

CI = 95 % significance level α = 5 % α = 0,05

z(c) for α = 0,05 from z-table is z(c) = 1,64

b) To calculate z(s) = ( p₁ - p ) / √ p₁*q₁ / n

z(s) = ( 0,71 - 0,70 )/ √ 0,71*0,29/616

z(s) = 0,01 /0,01828

z(s) = 0,547 ≈ 0,55

c) p-value for z(s) is from z-table p-value ≈ 0,7088

d) p-value > 0,05 then we accept H₀ we don´t have enough evidence to reject H₀

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3 years ago

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So,

#1: $n \geq 3 \frac{11}{16} + 4 \frac{1}{2}$

Convert to like improper fractions.
$n \geq \frac{59}{16} + \frac{72}{16}$

Add.
$n \geq \frac{131}{16}\ or\ 8 \frac{3}{16}$

So, one solution could be $8 \frac{3}{16}$ .

Another solution could by 9. There is also 10, 11, 12, etc., and all numbers in between.

#2: $k \ \textless \ 6 \frac{2}{5} * 15$

Convert into improper fraction form.
$k \ \textless \ \frac{32}{5} * 15$

Multiply.
$\frac{(2^5)(3)(5)}{5}$

Cross-cancel, and we have our final result.
$(2^5)(3) = 96$
k < 96

96 is not a solution.

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So is 94, 93, 92, etc, and all numbers in between.

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