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3 years ago

The measure of angle 1 is 30°, and the measure of angle 2 is 110°. na ofanale 12

Mathematics

1 answer:

LenaWriter [7]3 years ago

3 0

40 is the answer if your asking for the second angle

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10 POINTS

hoa [83]

It would be “going up a steady rate through the origin”

7 0

2 years ago

Oblicz granicę ciągu (pierwiastek z (n^2+6)/2n+2).

kap26 [50]

Answer:

umm what

Step-by-step explanation:

6 0

3 years ago

A successful basketball player has a height of 6 feet 9 inches, or 206 cm. Based on statistics from a data set, his height con

Sphinxa [80]

Answer:

The player's height is 4.45 standard deviations above the mean.

Step-by-step explanation:

Z-score:

Measures how many standard deviations a measure X is above or below the mean.

His height converts to the z score of 4.45.

This means that:

The player's height is 4.45 standard deviations above the mean.

8 0

3 years ago

Students in a high school mathematics class decided that their term project would be a study of

weeeeeb [17]

The parameter of interest in this scenario is the proportion of all 1,000 students within the school who have "strict" parents or guardians.

<h3>How to calculate a test statistic?</h3>

In Mathematics, the test statistics of a given sample is calculated by using this formula:

$Z=\frac{x\;-\;u}{\frac{\sigma}{\sqrt{n} } }$

<u>Where:</u>

x is the sample mean.

u is the mean.

is the standard deviation.

n is the sample size.

In conclusion, a possible test statistic which can be used to estimate this parameter is the proportion of all students who have "strict" parents or guardians, based on the collected sample.

Read more on test statistic here: brainly.com/question/4621112

#SPJ1

8 0

2 years ago

Let n be a natural number. Show that 3 | n3 −n

Vladimir [108]

Let $n=1$ . Then $n^3-n=1^3-1=0$ . By convention, every non-zero integer $n$ divides 0, so $3\vert n^3-n$ .

Suppose this relation holds for $n=k$ , i.e. $3\vert k^3-k$ . We then hope to show it must also hold for $n=k+1$ .

You have

$(k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)$

We assumed that $3\vert k^3-k$ , and it's clear that $3\vert 3(k^2+k)$ because $3(k^2+k)$ is a multiple of 3. This means the remainder upon divides $(k+1)^3-(k+1)$ must be 0, and therefore the relation holds for $n=k+1$ . This proves the statement.

4 0

3 years ago