How many fewer students voted for apple juice than milk apple juice was 19 milk was 28 what is the answer

A sphere and a cylinder have the same radius and height. The volume of the cylinder is 8 meters cubed. Yolanda found the volume

Mrrafil [7]

Answer:

Yolanda should have found the volume by multiplying 8 by four-thirds

Step-by-step explanation:

Here, we want to find Yolanda’s error.

we proceed as follows;

Mathematically, the volume of the cylinder can be calculated using the formula;

V = pi * r^2 * h

now since the radius of the cylinder and the height is equal, we can say that h = r

and thus, the volume can be written as

V = pi * r^2 * r = pi * r^3

Mathematically, the volume of a sphere can be calculated using the formula

V = 4/3 * pi * r^3

Now from the question we know that the value of pi * r^3 =8 which is the value for the volume of the cylinder.

Thus the volume of the sphere would be 4/3 * 8

so the correct answer here is B

5 0

2 years ago

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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 (n ≥ 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, n = 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:

H₀: The usual load is 13 credits, i.e. μ = 13.

Assume that the significance level of the test is, α = 0.05.

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

The (1 - α) % 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 z 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, μ = 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.

3 0

2 years ago

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When the sun is at a certain angle in the sky, a 50-foot building will cast a -foot shadow. What is the length of the shadow in

Serhud [2]

Answer:

0.40 foot

Step-by-step explanation:

In the first case, a 50 - foot building casts a- foot shadow (1-foot). Let the angle of elevation of the sun from the shadow be represented by θ.

Then:

Tan θ = $\frac{opposite}{adjacent}$

Tan θ = $\frac{50}{1}$

Tan θ = 50

⇒ θ = $Tan^{-1}$ 50

= 88.8542

= $88.85^{o}$

The angle of elevation is approximately $88.85^{o}$ .

For a 20-foot pole,

Tan θ = $\frac{opposite}{adjacent}$

Tan $88.85^{o}$ = $\frac{20}{x}$

x = $\frac{20}{Tan 88.85^{o} }$

= 0.4015

= 0.40 foot

The length of the shadow of the pole is 0.40 foot.

4 0

2 years ago

The proper way to write a single digit whole number in a sentence is:

Fynjy0 [20]

The correct answer for the given question above would be the last option. The proper way to write a single digit whole number if it is in a sentence should be SPELLED OUT and not written in numerals or roman numerals. For example, I have SIX apples; and not I have 6 apples.

6 0

2 years ago

Directions for questions 4 & 5: We selected a random sample of 100 StatCrunchU students, 67 females and 33 males, and analyz

GaryK [48]

Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender Mean Std. dev. n

Female 2577.75 1916.29 67

Male 3809.42 2379.47 33

We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

$M_d=M_1-M_2=2577.75-3809.42=-1231.67$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795$

The t-value for a 95% confidence interval is t=1.96.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.96 \cdot 475.795=932.54$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = -1231.67-932.54=-2164.21\\\\UL=M_d+t \cdot s_{M_d} = -1231.67+932.54=-299.13$

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

6 0

2 years ago