Match each data relationship with a type of correlation.

J+1/2=3 9/14 please show your work if possible . ty a) 1/7 b) 4 1/7 c) 1 5/9 d) 3 1/7

MatroZZZ [7]

J + 1/2= 3 9/14
convert 3 9/14 to improper fraction

J + 1/2= [(3*14)+9]/14

J + 1/2= 51/14
subtract 1/2 from both sides

J= 51/14 - 1/2
change both to common denominators (of 14 in this case)

J= 51/14 - 7/14
subtract numerators only

J= 44/14
simplify numerator and denominator by 2

J= 22/7
convert back to mixed number

J= 3 1/7

ANSWER: J= 3 1/7

Hope this helps! :)

6 0

4 years ago

1 + 4 = 5<br> 2 + 5 = 12<br> 3 + 6 = 21<br> 8 + 11 = ?

AnnZ [28]

The answer of 8+11=40

7 0

3 years ago

Read 2 more answers

10 snails is to be chosen from this population. Find the probability that the percentage of streaked-shelled snails in the sampl

Alex17521 [72]

Here is the full question:

$The \ shell \ of \ the \ land \ snail \ Limocolaria \ martensiana \ has \ two \ possible \ color$ $forms: \ streaked \ and \ pallid. \ In \ a \ certain \ population \ of \ these \ snails, \ 60\% \ o f \ the$ $individuals \ have \ streaked \ s hells.\ 16 \ Suppose \ that \ a \ random \ sample \ of \ 10 \ snails$ $is \ to \ be \ chosen \ from \ this \ population. \ F ind \ the \ probabilit y \ that \ the \ percentage \ of$

$streaked-shelled \ snails \ in \ t he \ sample \ will \ be$

$(a) 50\%. \ (b) 60\%. \ (c) 70\%.$

Answer:

(a) 0.2007

(b) 0.2510

(c) 0.2150

Step-by-step explanation:

Given that:

The sample size = 10

Sample proportion= 60% 0.6

Let X represents the no of streaked-shell snails.

$X \sim Binom (n =1 0, p = 0.60)$

The Probability mass function (X) is:

$P(X =x)= (^n_x) p^x(1-p)^{n-x}; x = 0,1,2,3...$

The Binomial probability with mean μ = np

= 10 * 0.6

= 6

Standard deviation σ = $\sqrt{np(1-p)}$

= $\sqrt{10*0.6*(1-0.6)}$

= 1.55

The probability that the percentage of the streaked-shelled snails in the sample will be:

a)

$P(X = 0.5) = ^nC_x p^x (1 -p) ^{(n-x)}$

$= ^{10}^C_5 * (0.6)^5(1-0.6)^{10-5}$

$= \dfrac{10!}{5!(10-5)!} * (0.6)^5(1-0.6)^{10-5}$

= 0.2007

b)

$P(X = 0.6) = ^nC_x p^x (1 -p) ^{(n-x)}$

$= ^{10}^C_6 * (0.6)^6(1-0.6)^{10-6}$

$= \dfrac{10!}{6!(10-6)!} * (0.6)^6(1-0.6)^{10-6}$

= 0.2510

c)

$P(X = 0.7) = ^nC_x p^x (1 -p) ^{(n-x)}$

$= ^{10}^C_7 * (0.6)^7(1-0.6)^{10-7}$

$= \dfrac{10!}{7!(10-7)!} * (0.6)^7(1-0.6)^{10-7}$

= 0.2150

7 0

3 years ago

In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some poin

FrozenT [24]

Answer: The length of segments between this point and the vertices of greater base are $7\frac{5}{7}$ and 18.

Step-by-step explanation:

Let ABCD is the trapezoid, ( shown in below diagram)

In which AB is the greater base and AB = 18 DC= 11, AD= 3 and BC = 7

Let P is the point where The extended legs meet,

So, according to the question, we have to find out : AP and BP

In Δ APB and Δ DPC,

∠ DPC ≅ ∠APB ( reflexive)

∠ PDC ≅ ∠ PAB ( By alternative interior angle theorem)

And, ∠ PCD ≅ ∠ PBA ( By alternative interior angle theorem)

Therefore, By AAA similarity postulate,

$\triangle APB\sim \triangle D PC$

Let, DP =x

⇒ $\frac{3+x}{18} = \frac{x}{11}$

⇒ 33 +11x = 18x

⇒ x = 33/7= $4\frac{5}{7}$

Thus, PD= $4\frac{5}{7}$

But, AP= PD + DA

AP= $4\frac{5}{7}+3 =7\frac{5}{7}$

Now, let PC =y,

⇒ $\frac{7+y}{18} = \frac{y}{11}$

⇒ 77 + 11y = 18y

⇒ y = 77/7 = 11

Thus, PC= 11

But, PB= PC + CB

PB= 11+7 = 18

7 0

3 years ago

Cindy woks Monday through Friday at a rate of $15.18 per hour. She takes a 1-hour lunch break. If she

Alex Ar [27]

11 hours-1
10*15.18
151.80*5
Answer is$ 759 is her total pay

7 0

3 years ago