Solve the equation. You will not get credit if you do not show your work or 4 points

How do you solve the inequality of -11.3 is equal to or greater than -4.7 + 4x

DerKrebs [107]

Answer: -1.65 ≥ x

Step-by-step explanation:

We usually express "equal to or greater than" by using the symbol ≥.

And we set up the inequality by looking at the word description, then we have:

-11.3 ≥ -4.7 + 4x plus 4.7 on both sides.

-6.6 ≥ 4x divide 4 on both sides.

-1.65 ≥ x

5 0

3 years ago

See image for question.

Ostrovityanka [42]

Answer:

See below

Step-by-step explanation:

#3

P(A) = 3/8
P(B) = 2/7

(i)

P(A and B) = P(A) * P(B) = 3/8*2/7 = 3/28

(ii)

P(A or B) = P(A) + P(B) - P(A and B) = 3/8 + 2/7 - 3/28 = 1/56(21 + 16 - 6) = 31/56

(iii)

P(not A and not B) = P(not A) * P(not B) = (1 - 3/8)*(1 - 2/7) = 5/8*5/7 = 25/56

<u>Another way:</u>

P(not A and not B) = 1 - P(A or B) = 1 - 31/56 = 25/56

#4

Outcomes with two fair coins: TT, TH, HT, HH

Outcomes with normal dice: 1 to 6

a)

P(odd number) = 3/6 = 1/2

b)

P(two heads and an even number) = 1/4*3/6 = 1/8

c)

P(head and tail) = 1/2
P(prime) = 3/6 = 1/2 (primes are 2, 3 and 5)
P(head, tail and prime) = 1/2*1/2 = 1/4

d)

P(two tails and odd number) = 1/4*3/6 = 1/8

4 0

3 years ago

Lily swims for 5 hours in stream that has a current of 1 mph. She leaves her dock, swims upstream for 2 miles and then swim back

antiseptic1488 [7]

Answer:

Swimming speed in still water is approximately 1.5 mph

Step-by-step explanation:

Given the speed of the current = 1 mph

distance she swims upstream =2 miles,

the total time = 5 hours.

She swam 2 miles upstream against the current and 2 miles back to the dock with the current. The formula that relates distance, time, and rate is $d = r t\ \ \ \ or \ \ \ t= \frac{d}{r}$

Let x be the speed in still water.

Then her speed with the current is x + 1, and

her speed against the current is x – 1.

Total time is equal to 2 miles with upstream and 2 miles downstream.

$\frac{2}{x+1}+\frac{2}{x-1}=5\\\\\frac{2(x-1)}{(x+1)(x-1)}+\frac{2(x+1)}{(x-1)(x+1)}=5\\\\\frac{2x-2}{(x^2-1)}+\frac{2x+2)}{(x^2-1)}=5\\\\\frac{2x-2+2x+2)}{(x^2-1)}=5\\\\4x=5(x^2-1)\\4x= 5x^2-5\\0= 5x^2-4x-5$

Now Using quadratic formula to solve above equation we get;

$x= \frac{-b \±\sqrt{b^2-4ac}} {2a}\\\\here \ \ a= 5,b=-4,c=-5\\\\x= \frac{-(-4) \±\sqrt{(-4)^2-4\times 5\times-5}} {2\times5}\\\\x=\frac{4 \±\sqrt{16+100}} {10}\\\\x=\frac{4 \±\sqrt{116}} {10}=x=\frac{4 \±2\sqrt{29}} {10}= x=\frac{2 \±\sqrt{29}} {5}\\ x \approx 1.5 \ or \ x = -0.7$

Since speed must be positive, Hence speed of still water is about 1.5 miles per hour.

3 0

3 years ago

Rápido y fácil, trigonometría. Urgente porfavor

enot [183]

Answer:

The angle $\theta$ measures approximately $28.81^o$

Step-by-step explanation:

Since this problem involves the finding of an angle in a right angle triangle for which we know the opposite side and the adjacent side, the trigonometric function that relates them is the tangent:

$tan(\theta)=\frac{opposite}{adjacent}$