Ask question

Ask question

All categories

3 years ago

11

A fruit bowl contains apples, oranges, and bananas. The radius of one of the oranges is 1.25 inches. What is the approximate vol

ume of the orange? Use 3.14 for . A. 1.02 cubic inches B. 19.63 cubic inches C. 8.18 cubic inches D. 6.54 cubic inches

1 answer:

Elden [556K]3 years ago

3 0

Answer:

8.18 cubic inches

Step-by-step explanation:

An orange is a sphere so you will use the formula for the volume of a sphere.

V= 4 /3 * π * r^3 r^3= 1.25* 1.25 * 1.25= 1.953

V= 4/3* 3.14 * 1.953 = 8.176=8.18

You might be interested in

HELP PLEASE WILL LEAVE GOOD REVIEW.

AleksandrR [38]

Answer:

Step-by-step explanation:

He needs 2.5 cm of ribbon left or in easier words 6 cm..-

5 0

2 years ago

What is the exact volume of the cone? with a 5 ft radius and a 16 ft height

andrew11 [14]

Answer:

The volume is 418.93 ft^3

Step-by-step explanation:

Here, we want to find the volume of a cone

Mathematically, we use the formula;

V = 1/3 * pi * r^2 * h

r = 5 ft

h = 16 ft

Substituting these values;

V = 1/3 * 3.142 * 5^2 * 16

V = 418.93 ft^3

3 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

Simplify the following expression and write in standard form: 3 x (6y - 4y + 5x) PLEASE HELP

mylen [45]

Answer: 15x^2+6xy

Step-by-step explanation: 3x(6y−4y+5x)

(3x)(6y+−4y+5x)

(3x)(6y)+(3x)(−4y)+(3x)(5x)

18xy−12xy+15x^2

15x^2+6xy

8 0

3 years ago

Please I need help. I need this ASAP, the screenshot is right below here.

Delicious77 [7]

Answer:

no idea

Step-by-step explanation:

no idea jsbdkanxlanvsksnsp

6 0

3 years ago