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2 years ago

N triangle xyz, m∠z > m∠x m∠y. which must be true about △xyz?

Mathematics

2 answers:

dmitriy555 [2]2 years ago

6 0

Answer:

The answer is D.

Step-by-step explanation:

liubo4ka [24]2 years ago

4 0

The choices are the below that can be found elsewhere:

m∠X + m∠Z < 90°

m∠Y > 90°

∠X and∠Y are complementary

m∠X + m∠Y < 90°

Since the given is m<Z > m<X +m<Y and <span>the sum of measure of angles of a triangle is equal 180 degrees so from this result that the last one choice need being true sure so m<X +m<Y < 90°</span>

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A class with n kids lines up for recess. The order in which the kids line up is random with eachordering being equally likely. T

Elis [28]

Answer:

A) P(Betty is first in line and mary is last) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

B) The method used is Relative frequency approach.

Step-by-step explanation:

From the question, we are told a sample of n kids line up for recess.

Now, the order in which they line up is random with each ordering being equally likely. Thus, this means that the probability of each kid to take a position is n(total of kids/positions).

Since we are being asked about 3 kids from the class, let's assign a letter to each kid:

J: John

B: Betty

M: Mary

A) Now, we want to find the probability that Betty is first in line or Mary is last in line.

In this case, the events are not mutually exclusive, since it's possible that "Betty is first but Mary is not last" or "Mary is last but Betty is not first" or "Betty is the first in line and Mary is last". Thus, there is an intersection between them and the probability is symbolized as;

P(B₁ ∪ Mₙ) = P(B₁) + P (Mₙ) - P(B₁ ∩ Mₙ) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

Where;

The suffix 1 refers to the first position while the suffix n refers to the last position.

Also, P(B₁ ∩ Mₙ) = P(B₁) × P(Mₙ/B₁)

This is because the events "Betty" and "Mary" are not independent since every time a kid takes his place the probability of the next one is affected.

B) The method used is Relative frequency approach.

In this method, the probabilities are usually assigned on the basis of experimentation or historical data.

For example, If A is an event we are considering, and we assume that we have performed the same experiment n times so that n is the number of times A could have occurred.

Also, let n_A be the number of times that A did occur.

Now, the relative frequency would be written as (n_A)/n.

Thus, in this method, we will define P(A) as:

P(A) = lim:n→∞[(n_A)/n]

7 0

2 years ago

To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assig

Keith_Richards [23]

Answer:

1. Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C$

2. D. 36

3. C. 34

4. B. 1.059

5. B. 8.02

Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Part 1

The hypothesis for this case are:

Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C$

Part 2

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have $p$ groups and on each group from $j=1,\dots,p$ we have $n_j$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}$

We need to find the mean for each group first and the grand mean.

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

If we apply the before formula we can find the mean for each group

$\bar X_A = 27$ , $\bar X_B = 24$ , $\bar X_C = 30$ . And the grand mean $\bar X = 27$

Now we can find the sum of squares between:

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

Each group have a sample size of 4 so then $n_j =4$

$SS_{between}=SS_{model}=4(27-27)^2 +4(24-27)^2 +4(30-27)^2=72$

The degrees of freedom for the variation Between is given by $df_{between}=k-1=3-1=2$ , Where k the number of groups k=3.

Now we can find the mean square between treatments (MSTR) we just need to use this formula:

$MSTR=\frac{SS_{between}}{k-1}=\frac{72}{2}=36$

D. 36

Part 3

For the mean square within treatments value first we need to find the sum of squares within and the degrees of freedom.

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

$SS_{error}=(20-27)^2 +(30-27)^2 +(25-27)^2 +(33-27)^2 +(22-24)^2 +(26-24)^2 +(20-24)^2 +(28-24)^2 +(40-30)^2 +(30-30)^2 +(28-30)^2 +(22-30)^2 =306$

And the degrees of freedom are given by:

$df_{within}=N-k =3*4 -3 = 12-3=9$ . N represent the total number of individuals we have 3 groups each one with a size of 4 individuals. And k the number of groups k=3.

And now we can find the mean square within treatments:

$MSE=\frac{SS_{within}}{N-k}=\frac{306}{9}=34$

C. 34

Part 4

The test statistic F is given by this formula:

$F=\frac{MSTR}{MSE}=\frac{36}{34}=1.059$

B. 1.059

Part 5

The critical value is from a F distribution with degrees of freedom in the numerator of 2 and on the denominator of 9 such that we have 0.01 of the area in the distribution on the right.

And we can use excel to find this critical value with this function:

"=F.INV(1-0.01,2,9)"

And we will see that the critical value is $F_{crit}=8.02$

B. 8.02

5 0

2 years ago

Complete the steps for solving 7 = –2x2 + 10x. Factor out of the variable terms. inside the parentheses and on the left side of

Mamont248 [21]

we have

$7=-2x^{2} +10x$

Factor the leading coefficient

$7=-2(x^{2} -5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$7-12.50=-2(x^{2} -5x+2.5^{2})$

$-5.50=-2(x^{2} -5x+2.5^{2})$

Divide both sides by $-2$

$2.75=(x^{2} -5x+2.5^{2})$

Rewrite as perfect squares

$2.75=(x-2.5)^{2}$

Taking the square roots of both sides (square root property of equality)

$x-2.5=(+/-)\sqrt{2.75}$

Remember that

$\sqrt{2.75}=\sqrt{\frac{11}{4}}= \frac{\sqrt{11}}{2}$

$x-2.5=(+/-)\frac{\sqrt{11}}{2}$

$x=2.5(+/-)\frac{\sqrt{11}}{2}$

$x=2.5+\frac{\sqrt{11}}{2}=\frac{5+\sqrt{11}}{2}$

$x=2.5-\frac{\sqrt{11}}{2}=\frac{5-\sqrt{11}}{2}$

<u>the answer is</u>

The solutions are

$x=\frac{5+\sqrt{11}}{2}$

$x=\frac{5-\sqrt{11}}{2}$

5 0

2 years ago

Read 2 more answers

There are 35 students in art class and 57 students in dance class. Find the number of students who are either in art class or in

Murrr4er [49]

Answer:

a. 80 students

b. 92 students

Step-by-step explanation:

Represent arts students with A and Dance students with D.

So, we have,

n(A) = 35

n(D) = 57

Required

Determine n(A or D)

Solving (a):

Here, we have:

n(A and D) = 12

n(A or D) is calculated as thus:

n(A or D) = n(A) + n(D) - n(A and D)

n(A or D) = 35 + 57 - 12

n(A or D) = 80

b. From the given details

n(A and D) = 0 because both students are not mixed up as in (a) above

Using the same formula as (a).

n(A or D) = n(A) + n(D) - n(A and D)

n(A or D) = 35 + 57 - 0

n(A or D) = 92

8 0

2 years ago

Given: ∠BCD is right; BC ≅ DC; DF ≅ BF; FA ≅ FE Triangles A C D and E C B overlap and intersect at point F. Point B of triangle

Aloiza [94]

Answer:

1) ΔCBF ≅ ΔCDF by (SSS)

2) ΔBFA ≅ ΔDFE by (SAS)

3) ΔCBE ≅ ΔCDA by (HL)

Step-by-step explanation:

1) Since BC ≅ DC and DF ≅ BF where CF ≅ CF (reflective property) we have;

ΔCBF ≅ ΔCDF by Side Side Side (SSS) rule of congruency

2) Since DF ≅ BF and FA ≅ FE where ∠DFE = ∠BFA (alternate angles)

Therefore;

ΔBFA ≅ ΔDFE by Side Angle Side (SAS) rule of congruency

3) Since FA ≅ FE and DF ≅ BF then where EB = FE + BF and AD = FA + DF

Where:

EB and AD are the hypotenuse sides of ΔCBE and ΔCDA respectively

We have that;

EB = AD from FE + BF = FA + DF

Where we also have BC ≅ DC

Where:

BC and DC are the legs of ΔCBE and ΔCDA respectively

Then we have the following relation;

ΔCBE ≅ ΔCDA by Hypotenuse Leg (HL).

4 0

2 years ago

Read 2 more answers