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Fynjy0 [20]
2 years ago
10

Which numbered choice shows a set of numbers whose product is -20 and whose sum is +1?

Mathematics
1 answer:
Ira Lisetskai [31]2 years ago
5 0
5 and -4 are the numbers
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Please help im so confused at math pls help
exis [7]

Answer:

where is the question ??

6 0
2 years ago
What is the surface area of the cylinder with height 7 cm and radius 5 cm? Round
nalin [4]

Answer:

376.991 cm²

Step-by-step explanation:

SA_{cylinder} = 2\pi r^{2}  + 2\pi rh

Surface Area = 2π(5²) + 2π(5)(7)

                      = 50π + 70π

                      = 120π

                      = 376.9911184... ≈ 376.991

<h2>Derive Surface Area of A Cylinder</h2>

If you take a look at a net of a cylinder, you can see it is composed of two circles(bases) and a rectangular strip with the length of the diameter of the circle

The formula for the area of a circle = πr², and since there are 2, it is 2πr²

The formula for circumference of a circle = 2πr, and since we are multiplying that by the height by the height of the cylinder, it is 2πrh

∴ 2πr² + 2πrh

5 0
2 years ago
PLEASE SOLVE FOR ME &lt;3
natta225 [31]

Step-by-step explanation:

At the dance, soft drinks were 4 for $2.20 this use relationship to fill in the table.

A.) fill in the table:

# of drinks: 1 2 3 4

Princes ($) | | | | |

B.) why is the constant of proportionality?

C.) if you were to buy 22 soft drinks, how much would it cost?

5 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
Which letter will correspond to the location of the following points when plotted on the number line below?
Alinara [238K]

Answer:

A, E, B, D, C

Step-by-step explanation:

1) 6 is A

2) √83 = 9.11, so it is E

3) √40 = 6.3, so it is B

4) √64 = 8, so it is D

5) 7.5 is C

5 0
2 years ago
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