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Art [367]
2 years ago
7

if 1000 spherical of radius 1 cm are melted to form a bigger bulb. Then find the radius of bigger bulb​

Mathematics
1 answer:
never [62]2 years ago
3 0

Answer:

10 cm

Step-by-step explanation:

Given:

No. of small spherical bulb = 1,000

radius (r) of smaller bulbs = 1 cm

Required:

radius of the bigger bulb

SOLUTION:

The following equation represents the relationship of the volume of the smaller and bigger bulb,

\frac{V_2}{V_1} = 1,000

Where,

V_2 = volume of bigger bulb

V_1 = volume of smaller bulb

1,000 is the number of smaller bulbs melted to form the bigger bulb.

Volume of a sphere is given as, ⁴/3πr³

Therefore:

V_2 = ⁴/3*π*r³ = 4πr³/3

V_1 = ⁴/3*πr³ = ⁴/3*π*(1)³ = ⁴/3π*1 = 4π/3

Plug the above values into the equation below:

\frac{V_2}{V_1} = 1,000

\frac{\frac{4*pie*r^3}{3}}{\frac{4*pie}{3}} = 1,000

\frac{4*pie*r^3}{3}*{\frac{3}{4*pie} = 1,000

\frac{4*pie*r^3*3}{3*4*pie} = 1,000

\frac{12*pie*r^3}{12*pie} = 1,000

r^3 = 1,000 (12pie cancels 12 pie)

r = 10 (taking the cube root of each side)

Radius of the bigger bulb = 10 cm

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A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec
Ket [755]

Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18

The sample size is, <em>n</em> = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

(a)

The null hypothesis is:

<em>H</em>₀: The usual load is 13 credits, i.e. <em>μ</em> = 13.

Assume that the significance level of the test is, <em>α</em> = 0.05.

Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.

The (1 - <em>α</em>) % confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\times SE

For 5% level of significance the two tailed critical value of <em>z</em> is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct the 95% confidence interval as follows:

CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)

As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

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2 years ago
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Answer:

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Step-by-step explanation:

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