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2 years ago

Write a statement

Mathematics

1 answer:

nekit [7.7K]2 years ago

6 0

9514 1404 393

Answer:

C) ΔDCE ≅ ΔDQR

Step-by-step explanation:

Corresponding vertices can be identified by the number of arcs signifying congruence.

1 mark: angles D and D

2 marks: angles C and Q

3 marks: angles E and R

Corresponding angles are listed in the same order in the congruence statement:

ΔDCE ≅ ΔDQR

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What is the mean absolute deviation for 7, 5, 1, 6, 2, and 3?

Gwar [14]

7, 5, 1, 6, 2 Mean Absolute Deviation = 2.16

3, 6, 3, 2, 1 Mean Absolute Deviation = 1.2

3 0

2 years ago

Write an equation in slope-intercept form for the line that has a slope of -4/5 and passes through (5, 10)

garik1379 [7]

Let m = slope = -4/5
P(x,y) = P(5,10)

Slope-intercept form for the equation of a line:
y=mx+c

Plug m= -4/5 into the slope-intercept formula
y = -4/5x + c

Plug P(5,10) into point-slope formula
y = -4/5x + c
10 = -4/5(5) + c
10 = -4 + c
substract -4 from both sides,
10 - (-4) = -4 - (-4) + c
14 = c

So, the equation is y = -4/5x + 14

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2 years ago

Brainliest if correct

kari74 [83]

Answer:

A.1/3

Step-by-step explanation:

1 Pick two points on the line and determine their coordinates.

2 Determine the difference in y-coordinates of these two points (rise).

3 Determine the difference in x-coordinates for these two points (run).

4 Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).

7 0

3 years ago

You could win a giant jar of M&Ms, if only you could guess how many are in the jar. You find a smallerjar such that the gian

sleet_krkn [62]

Answer:

There are 7360 M&Ms in larger jar.

Step-by-step explanation:

We are given the following in the question:

A giant jar is such that it is exactly four times times the smaller jar in dimensions.

If r and h is the radius and height of the cylindrical jar respectively, then the dimensions of the larger are:

$R = 4r\\H=4h$

Number of M&Ms in smaller jar = 115

Volume of smaller jar =

$v = \dfrac{1}{3}\pi r^2 h$

Volume of larger jar =

$V =\dfrac{1}{3}\pi R^2 H=\dfrac{64}{3}\pi r^2 h$

Number of M&Ms in larger jar:

$v = \dfrac{1}{3}\pi r^2 h\rightarrow 115\\\\V=\dfrac{64}{3}\pi r^2 h\rightarrow 115\times 64\\\\V=\dfrac{64}{3}\pi r^2 h\rightarrow 7360$

Thus, there are 7360 M&Ms in larger jar.

6 0

2 years ago

A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January

Blababa [14]

Answer:

a) 27.89% probability that two have ever boycotted goods for ethical reasons

b) 41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) 41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) The expected number is 2.3 and the standard deviation is 1.33.

Step-by-step explanation:

We use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

23% of the respondents have boycotted goods for ethical reasons.

This means that $p = 0.23$

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

This is P(X = 2) when n = 6. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789$

27.89% probability that two have ever boycotted goods for ethical reasons

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Either less than two have, or at least two. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084$

$P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181$

41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

Now n = 10.

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,5}.(0.23)^{5}.(0.77)^{5} = 0.0439$

$P(X = 6) = C_{10,6}.(0.23)^{6}.(0.77)^{4} = 0.0109$

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2343 + 0.1225 + 0.0439 + 0.0109 = 0.4116$

41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

$E(X) = np = 10*0.23 = 2.3$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.23*0.77} = 1.33$

The expected number is 2.3 and the standard deviation is 1.33.

5 0

2 years ago