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3 years ago

The sum of two numbers is 17 and their difference is ?

1 answer:

3 0

Impossible to tell. there is an infinite number of solutions to x+y=17, without knowing the difference of the numbers we cannot solve for X and Y

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A physical education department wanted a single test of upper body strength that was easy to administer. Dips on the parallel ba

Gennadij [26K]

Answer:

24.23 ; 13.47 ; 0.782 ; p value < 0.00001 ;7.45

Step-by-step explanation:

Given that :

ΣX = 3,416, ΣY = 1,899, SDX = 8.84, SDY = 4.70, ΣZXZY = 109.416

Sample size, N = 141

Mean of X:

£X / N = 3416 / 141 = 24.23

Mean of Y:

£Y / N = 1899 / 141 = 13.47

B.) Correlation between the 2 variables :

£ZXZY / N - 1

= 109.416 / 140

= 0.782

C.)

Level of significance and p value :

T = [Rsqrt(n-2)] ÷ sqrt(1 - R²)

T = [0.782 * sqrt(139)] ÷ sqrt(1-0.782^2)

T = 14.79

P value from Tscore calculator

Df = 141 - 2 = 139 ; α = 0.05

Pvalue = 0.00001

Pvalue < α ; Hence, a significant positive relationship exists.

To obtain regression model :

Slope = £ZXZY / SDX²

m = 109.416 / 8.84^2 = 1. 4

yintercept = meanY - slope*x

yintercept = 13.47 - 1.4*24.3

yintercept = - 20.55

Equation in slope intercept form:

y = mx + c

y = 1.4x - 20.55

For x = 20

y = 1.4(20) - 20.55

y = 7.45

5 0

2 years ago

A standard deck of cards contains 52 cards. One card is selected from the deck.(a)Compute the probability of randomly selecting

Aneli [31]

a. There are four 5s that can be drawn, and $\binom43=4$ ways of drawing any three of them. There are $\binom{52}3=22,100$ ways of drawing any three cards from the deck. So the probability of drawing three 5s is

$\dfrac{\binom43}{\binom{52}3}=\dfrac4{22,100}=\dfrac1{5525}\approx0.00018$

In case you're asked about the probability of drawing a 3 or a 5 (and NOT three 5s), then there are 8 possible cards (four each of 3 and 5) that interest you, with a probability of $\frac8{52}=\frac2{13}\approx0.15$ of getting drawn.

b. Similar to the second case considered in part (a), there are now 12 cards of interest with a probability $\frac{12}{52}=\frac3{13}\approx0.23$ of being drawn.

c. There are four 6s in the deck, and thirteen diamonds, one of which is a 6. That makes 4 + 13 - 1 = 16 cards of interest (subtract 1 because the 6 of diamonds is being double counted by the 4 and 13), hence a probability of $\frac{16}{52}=\frac4{13}\approx0.31$ .

- - -

Note: $\binom nk$ is the binomial coefficient,

$\dbinom nk=\dfrac{n!}{k!(n-k)!}={}_nC_k=C(n,k)=n\text{ choose }k$

6 0

3 years ago

Figure A is translated 3 units right and 2 units up. The translated figure is labeled figure B. Figure B is reflected over the x

mojhsa [17]

Answer:

A Is congruent to B and B Is congruent to C

Step-by-step explanation:

Given Figures A, B and C on a coordinate plane with the coordinates below:

Triangle A has points (1, -2), (3, -2), (3, -5).

Triangle B has points (4, 0), (6, 0), (6, -3).

Triangle C has points (4, 0), (6, 0), (6, 3).

The fact that each triangle is a right triangle only shows that the triangles are similar.

For any two shapes to be congruent, they must have in addition to the same shape, the same size.

Therefore, for Triangle A to be congruent to Triangle C, Triangle A must be congruent to Triangle B and Triangle B must be congruent to Triangle C.

7 0

2 years ago

Given: m∠A = m∠C = 90° AB ║ DC Prove: ∆ABD ≅ ∆CBD

Genrish500 [490]

Answer:

Angle-Angle-Side (AAS)

Step-by-step explanation:

Given triangles; ΔABD and ΔCBD

<A = <C = $90^{0}$ (right angle property)

AB ║ DC (given)

Then,

DA ║ CB

<B ≅ <D (congruent property)

DB ≅ BD (similarity property)

<D + <B = <B + <D (complementary angles, and property of triangles)

Therefore by Angle-Angle-Side (AAS),

∆ABD ≅ ∆CBD

4 0

2 years ago

Plz help on 4,8 really need answers to get some point

mafiozo [28]

4) The line segments formed are parallel because they never meet.

8) The answer is B. That is because there is a right angle and right angles are formed by perpendicular lines.

Hope this helps :)<span />

5 0

3 years ago