We have
$\overline {UV} = 16.8 cm\\\\\overline {WV} = 7.9 cm\\\\ Since all lines are scalars, \overline{WV} = \overline{VW}\\\textrm{We see that \overline{UW} + \overline{WV} = \overline{UV}\\\\\overline {UW} \\ = \overline {UV} - \overline {WV} = 16.9 - 7.9 = 8.9 cm}$

7 0

1 year ago

Find two numbers, if their sum is 49 and their difference is 1/2 .

amm1812

Answer:

$x=24\frac{3}{4}\\ y=24\frac{1}{4}$

Step-by-step explanation:

Let the two numbers be x and y

So we have:

$x +y = 49\\x-y=\frac{1}{2}$

using the elimination method, add the two equations together:

$2x=49\frac{1}{2} \\2x=\frac{99}{2}$

solve for x:

$x=\frac{99}{4}\\x=24\frac{3}{4}$

substitute our value for x back into the first equation we made:

$x+y=49\\(\frac{99}{4}) + y=49\\y=49-\frac{99}{4}\\y=\frac{97}{4} \\ y=24\frac{1}{4}$

therefore,

$x=24\frac{3}{4}\\ y=24\frac{1}{4}$

6 0

2 years ago

Before 1918, approximately 40% of the wolves in a region were male, and 60% were female. However, cattle ranchers in this area h

Goshia [24]

Answer:

P(≥ 7 males) = 0.0548

Step-by-step explanation:

This is a binomial probability distribution problem.

We are told that Before 1918;

P(male) = 40% = 0.4

P(female) = 60% = 0.6

n = 10

Thus;probability that 7 or more were male is;

P(≥ 7 males) = P(7) + P(8) + P(9) + P(10)

Now, binomial probability formula is;

P(x) = [n!/((n - x)! × x!)] × p^(x) × q^(n - x)

Now, p = 0.4 and q = 0.6.

Also, n = 10

Thus;

P(7) = [10!/((10 - 7)! × 7!)] × 0.4^(7) × 0.6^(10 - 7)

P(7) = 0.0425

P(8) = [10!/((10 - 8)! × 8!)] × 0.4^(8) × 0.6^(10 - 8)

P(8) = 0.0106

P(9) = [10!/((10 - 9)! × 9!)] × 0.4^(9) × 0.6^(10 - 9)

P(9) = 0.0016

P(10) = [10!/((10 - 10)! × 10!)] × 0.4^(10) × 0.6^(10 - 10)

P(10) = 0.0001

Thus;

P(≥ 7 males) = 0.0425 + 0.0106 + 0.0016 + 0.0001 = 0.0548

3 0

2 years ago

Evaluate 2/3x -16 when x= 18A.-6B.-4C.2D.4​

Evaluate 2/3x -16 when x= 18A.-6B.-4C.2D.4