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Kipish [7]
3 years ago
5

Which side lengths form a right triangle ?

Mathematics
1 answer:
True [87]3 years ago
4 0

Answer:

B

Step-by-step explanation:

1^2 + 8^2 = c^2

1 + 64 = c^2

65 = c^2

√c^2 = √65

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There are a total of 15 bicycles and tricycles.There are 37 wheels all together if we count them up. How many tricycles and how
m_a_m_a [10]

Answer: there are eight bicycles and 7 tricycles.

Step-by-step explanation:

Let x represent the number of bicycles that are there.

Let y represent the number of tricycles that are there.

There are a total of 15 bicycles and tricycles. This means that

x + y = 15

A bicycle has 2 wheels and a tricycle has 3 wheels. There are 37 wheels all together if we count them up. This means that

2x + 3y = 37- - - - - - - - - - - - - 1

Substituting x = 15 - y into equation 1, it becomes

2(15 - y) + 3y = 37

30 - 2y + 3y = 37

- 2y + 3y = 37 - 30

y = 7

x = 15 - y = 15 - 7

x = 8

7 0
3 years ago
The time required for workers to produce each unit of a product decreases as the workers become more familiar with the productio
Marat540 [252]

Answer: 2.79 hours.

Step-by-step explanation:

Given that the function for the learning process is T(x) = 2 + 0.3 1 x , where T(x) is the time, in hours, required to produce the xth unit

To calculate the time for the new worker to produce 10 units, substitute 10 for x in the equation above.

T(x) = 2 + 0.31 (10)

T(x) = 2 + 3.1

T(x) = 5.1 hours

To calculate the time for the new worker to produce 19 units, substitute 19 for x in the equation above.

T(x) = 2 + 0.31(19)

T(x) = 2 + 5.89

T(x) = 7.89 hours

The time required for a new worker to produce units 10 through 19 will be

7.89 - 5.1 = 2.79 hours

7 0
3 years ago
Write the equation of the line that passes through the given points. Include your work in your final answer. Type your answer in
Harman [31]
The equation you use for the slope of this kind of problem is (y2-y1)/(x2-x1). x1 is the first X coordinate, x2 is the second, y1 is the first y coordinate, y2 is the second.

(5-3)/(2-2)
2/0
This equation has an UNDEFINED SLOPE, as it is just a vertical line.
3 0
4 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
WILL GIVE BRAINLIEST IF YOU HELP ME
Oduvanchick [21]
The answer is D. point S
8 0
3 years ago
Read 2 more answers
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