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Tom [10]
3 years ago
6

P(2,4),parallel to y=-3×+4

Mathematics
1 answer:
daser333 [38]3 years ago
5 0
4 = - 3(2) + c
4 = - 6 + c
c = 10
y = - 3x + 10
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On average, an American hummingbird flaps its wings about 3,180 times per minute.
Elden [556K]

Answer:

53 times per second

Step-by-step explanation:

There are 60 seconds in 1 minute, so divide the 3180 flaps per minute into 60 equal parts.

3180 / 60 = 53

5 0
2 years ago
I NEED HELP PLSSS<br><br> Find the value of x.
tankabanditka [31]

Answer:

1) x= 43

2) x= 54

Step-by-step explanation:

you add the two angles given, then take that answer and subtract it from 180!

i hope this helped!!

7 0
2 years ago
Read 2 more answers
a major metroplitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. they aske
Arisa [49]

Answer:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 1600, \pi = 0.4

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.3685

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.4315

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

7 0
2 years ago
36 is 0.09% of what number
alexgriva [62]
Х-100%
36-0,06%
x=(36*100)\0,06=60000
Answer:60000
8 0
2 years ago
The following graphs have a scale assigned to them: The area of each grid
dezoksy [38]

The graphs that are density curves for a continuous random variable are: Graph A, C, D and E.

<h3>How to determine the density curves?</h3>

In Geometry, the area of the density curves for a continuous random variable must always be equal to one (1). Thus, we would test this rule in each of the curves:

Area A = (1 × 5 + 1 × 3 + 1 × 2) × 0.1

Area A = 10 × 0.1

Area A = 1 sq. units (True).

For curve B, we have:

Area B = (3 × 3) × 0.1

Area B = 9 × 0.1

Area B = 0.9 sq. units (False).

For curve C, we have:

Area C = (3 × 4 - 2 × 1) × 0.1

Area C = 10 × 0.1

Area C = 1 sq. units (False).

For curve D, we have:

Area D = (1 × 4 + 1 × 3 + 1 × 2 + 1 × 1) × 0.1

Area D = 10 × 0.1

Area D = 1 sq. units (True).

For curve E, we have:

Area E = (1/2 × 4 × 5) × 0.1

Area E = 10 × 0.1

Area E = 1 sq. units (True).

Read more on density curves here: brainly.com/question/26559908

#SPJ1

6 0
2 years ago
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