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3 years ago

Can someone please clearly and correctly work out Q19? Would appreciate it a lot! :)

Mathematics

1 answer:

DiKsa [7]3 years ago

6 0

$x^{2} +10x+30$
I always start by moving the last term to the other side
$x^{2} +10x=-30$
To complete the square you take term 'b' , divide it by 2 then square it
$( \frac{10}{2} )^2$ = 25
We have to add 25 to what we have and what you do to one side you have to do to the other
$x^{2} +10x+25=-30+25$
$x^{2} +10x+25=-5$
Now we factor what is on the left side of the equation
$(x+5)^2=-5$
The last step is to move our right side of the equation back over
$(x+5)^2+5$ - which is your answer :)))

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2 years ago

For the problem -4(-5)(9), will the answer be positive or negative?

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2 years ago

Convert the measurement as indicated. 5.2 mi to feet

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2 years ago

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Rus_ich [418]

Answer:

1287

Step-by-step explanation:

Number of people in elevator , n = 8

Number of floors, r = 6

Since they all look alike, arrangement doesn't matter, the we use combination :

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2 years ago

Soft drink consumption in New Zealand. A survey commissioned by the Southern Cross Healthcare Group reported that 16% of New Zea

lidiya [134]

<h2>Answer with explanation:</h2>

Given : The proportion of New Zealanders consume five or more servings of soft drinks per week : $\hat{p}= 0.16$

a) The number of survey respondents reported that they consume five or more servings of soft drinks per week = 2006

b) Confidence interval for population proportion (p) :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ = Sample proportion.

n= Sample size.

z* = Critical value.

For n= 2006 , $\hat{p}= 0.16$ and critical value for 95% confidence interval : z* = 1.96

Then , the required confidence interval will be :

$0.16\pm (1.96)\sqrt{\dfrac{ 0.16(1- 0.16)}{2006}}$

$0.16\pm (1.96)\sqrt{ 0.000066999002991}$

$0.16\pm (1.96)(0.00818529186963)$

$0.16\pm 0.0160431720645$

$(0.16-0.0160431720645,\ 0.16+0.0160431720645)$

$(0.143956827935,\ 0.176043172064)\approx(0.144,\ 0.176)$

i.e. A 95% confidence interval for the proportion of New Zealanders who report that they consume five or more servings of soft drinks per week. = $(0.144,\ 0.176)$

c) $(0.144,\ 0.176)=(0.144\times100\%,\ 0.176\times100\%)$

$=(14.4\%,\ 17.6\%)$ 176\times100\%)[/tex]

d) The estimate might be biased because the survey is taken online that means offline New Zealanders are out of consideration.

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3 years ago