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3 years ago

I need help please help

Mathematics

1 answer:

pishuonlain [190]3 years ago

5 0

G 3/4 hope this helps

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Factorise the following 15xy - 24x squared please

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3x(5y-8) hope this helps

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John earns $300 for a 40-hour week. If he receives time and a half for overtime, what is his hourly overtime wage? A.$7.50 B. $9

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Answer: A.$7.50

Step-by-step explanation:

$300 divided by 40 = $7.50

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What is the zero of the linear function?

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Answer:

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(-6,-2) (-2,6) (6,2) (-2,-6)

LuckyWell [14K]

Answer:

(-6,-2)

Step-by-step explanation:

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I think that was what you were asking. I hope.

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3 years ago

A botanist wishes to estimate the mean number of seeds for a certain fruit. She samples 11 specimens and finds the average numbe

Aliun [14]

Answer:

a)The 98% of confidence intervals are

Lower bound of CI = 47 -3.82436 = 43.1756

Upper bound of CI = 47 + 3.82436 = 50.8243

b) The conditions are required for the validity of the interval

A) μ known

Step-by-step explanation:

Explanation:-

The given sample size is 'n' =11

Given the average number of seeds is 47 with a standard deviation of 7

Sample mean (x⁻) = 47

Standard deviation (S) = 7

The 98% of confidence intervals are

$(x^{-} - t_{0.02} \frac{S}{\sqrt{n} } , x^{-} + t_{0.02}\frac{S}{\sqrt{n} } )$

$(47 - t_{0.02} \frac{7}{\sqrt{11} } , 47+ t_{0.02}\frac{7}{\sqrt{11} } )$

The degrees of freedom = n-1 =11-1 =10

t₀.₀₂= 1.812 ( from t - table)

now the intervals are

$(47 - 1.812 \frac{7}{\sqrt{11} } , 47+ 1.812\frac{7}{\sqrt{11} } )$

Lower bound of CI = 47 -3.82436 = 43.1756

Upper bound of CI = 47 + 3.82436 = 50.8243

The conditions are required for the validity of the interval

A) μ known

Explanation:-

If a random sample xi of size 'n' has been drawn from a normal population with a specified mean (μ) known.

The limit for (μ) is given by

$(x^{-} - t_{0.02} \frac{S}{\sqrt{n} } , x^{-} + t_{0.02}\frac{S}{\sqrt{n} } )$

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3 years ago