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dimaraw [331]
3 years ago
6

Which graph summarize a set of data using an interval scale?

Mathematics
1 answer:
nignag [31]3 years ago
8 0
To summarize a set of data using an interval scale, you would use a box plot.
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Question 14 (1 point)
valina [46]

Answer:

C(x) = 25 + 2x

Step-by-step explanation:

He already has 25 cards.

He is collecting 2 cards per month. In x months, he will have collected:

2 * x = 2x cards

That means that the number of cards he will have after x months will be the sum of the cards he already has and the cards he collects per month:

C(x) = 25 + 2x

8 0
3 years ago
Write an inequality for the graph
natima [27]
The open circle means the inequality will be greater than or equal to (≥) or less than or equal to (≤).
A closed circle means the inequality will be greater than (>) or less than (<)

An arrow pointing right to increasingly positive values means the inequality is getting greater (> or ≥)
A narrowing pointing left to increasingly negative values means the inequality is getting lesser (< or ≤)

So for this graph with an open circle and rightward pointing arrow, “x” will be some number on the number line greater than the first point of -38:
x > -38
6 0
3 years ago
What is the arc length of an arc with radius 10 in and central angle 60 degrees ? Show your work.
emmasim [6.3K]

Answer:

\huge \boxed{ \boxed{ \red{{s \approx \: 10.47}}}}

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • geometry
  • PEMDAS
<h3>tips and formulas:</h3>
  • s =  \frac{\pi {r} \theta}{180}
<h3>given:</h3>
  • r=10
  • \theta = 60
<h3>let's solve:</h3>
  1. \sf sustitute \: the \: value s \: of \:  \theta \: and \: r :  \\  =\frac{\pi \times 10 \times 60}{180}
  2. \sf rediuce \: 60 :  \\  =\frac{\pi \times 10 \times  \cancel{60}  \: ^{1} }{ \cancel{180}  \:  \:^{3} }  \\  =\frac{\pi \times 10}{3}
  3. \sf use \: 3.14 \: for \: \pi :  \\ = \frac{3.14 \times 10}{3}
  4. \sf simplify \: multipication :  \\  =  \frac{31.4}{3}  \\  \approx \: 10.47
5 0
3 years ago
How do i calculate percent accuracy? I got 20 questions out of 21 correct but I need to figure out the percent accuracy for anot
Mandarinka [93]
You need to do 20 divide by 21 to get a decimal. Then if you multiply the decimal by hundred you should get the percentage.
7 0
3 years ago
While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem
Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, H_0 : p = 0.013      {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, H_A : p > 0.013      {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion faulty modems= \frac{10}{367} = 0.027

           n = sample of modems = 367

So, <u><em>the test statistics</em></u>  =  \frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }

                                     =  2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0
3 years ago
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