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Phantasy [73]
3 years ago
15

Please help me now plz I will mark you brainliest

Mathematics
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

On the circle

Step-by-step explanation:

Because all of the points on a circle are equidistant from the center, if a circle has a radius of 10 then all of the points are 10 units from the center. Using the Pythagorean Theorem, you know that the distance from (4,0) to (-2,8) is:

\sqrt{(4-(-2))^2+(8-0)^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10, meaning that it is on the circle. Hope this helps!

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Y_Kistochka [10]

Answer:

8. 24

9.55.3

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that
ICE Princess25 [194]

Answer:

The probability that event B will occur is 0.45

Step-by-step explanation:

Given;

probability that event A occurs, P(A) = 0.4

the probability that events A and B both occur, P(A ∩ B) = 0.25

the probability that either event A or event B occurs, P(A ∪ B) = 0.6

To determine the probability that event B will occur, we use probability addition rule;

P(A) + P(B) = P(A ∩ B) + P(A ∪ B)

0.4 + P(B) = 0.25 + 0.6

0.4 + P(B) = 0.85

P(B) = 0.85 - 0.4

P(B) = 0.45

Therefore, the probability that event B will occur is 0.45

4 0
3 years ago
Divide use partial quotients 28 divided by 514
ella [17]
28÷514=0.
514÷28=18.
Partial Quotients
6 0
3 years ago
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0
3 years ago
What is the simplified form of the expression 0.3 (4x - 4y)?
Agata [3.3K]

Answer:

1.2x-1.2y

Step-by-step explanation:

8 0
3 years ago
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