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Stels [109]
3 years ago
5

A cone has radius 5x cm and a height 12x cm.A sphere has radius r cm.The cone has the same total surface area as the sphere.Show

that r²=45÷2 x². (The curved surface area,A, of a cone with radius r and slant height l is A=πrl.) (The surface area,A, of a sphere with radius r is A=4πr².)
someone please help with this last one
Mathematics
1 answer:
Lana71 [14]3 years ago
8 0

Answer:

r^2 = 45÷2 x^2

Step-by-step explanation:

Details of cone:

Radius(r) = 5x cm

Height(h) = 12x cm

Slant height(l) = √(12x^2 + 5x^2)

l = 13x cm

Total surface area of a cone (T. S. A) = πrl + πr²

T. S. A = π * 5x * 13x + π5x^2

T.S.A = π65x^2 + π5x^2

T.S.A = π65x^2 + π25x^2

T.S.A = 90πx^2 - - - - - (1)

Details of sphere:

Radius (r) = r cm

Total surface area = 4πr²

Total Surface area of sphere= 4πr²cm - - - (2)

Equating (1) and (2) to calculate r

4πr² = 90πx^2

r^2 = 90πx^2 / 4π

r^2 = 90π (x^2) / 4π

r^2 = 45/2 (x^2) ;

r^2 = 45÷2 x^2

Hence the proof

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At Ajax Spring Water, a half-liter bottle of soft drink is supposed to contain a mean of 519 ml. The filling process follows a n
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Answer:

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2) C. H0: mu = 519, H_1: mu not equal to 519, reject if z> 1.96 or z< -1.96.

3) Yes. There is enough evidence to support the claim that the true mean is not 519 ml.

Step-by-step explanation:

1) When the population follows a normal distribution, it is correct to assume a normal distribution for the sample mean.

2) As it is a two-tailed decision rule, we are interested in detecting a significant difference below and above the mean. This is why we use the unequal sign in the alternative hypothesis.

The null hypothesis state that there is not significant difference from 519.

The critical value for a significance level of 5% is z=1.96.

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3) The claim is that the true mean is not 519 ml.

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H_0: \mu=519\\\\H_a:\mu\neq 519

The significance level is 0.05.

The sample has a size n=16.

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We can calculate the standard error as:

\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{16}}=1.5

Then, we can calculate the z-statistic as:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{522-519}{1.5}=\dfrac{3}{1.5}=2

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As the P-value (0.046) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the true mean is not 519 ml.

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