If Candler and Oteen are 312 inches apart on the map, what is the actual distance between Candler and Oteen in miles?

A shoe store charges a retail price of $39 for a certain type of sneaker. This retail price is 30% more than the amount it costs

Vedmedyk [2.9K]

The answer is $31.20

Hope this helped :)

4 0

3 years ago

Double Points! (i have no clue how to do this) Find the value of the variables in the figure.

Aleks [24]

Answer:

x=12 and y=31

Step-by-step explanation:

First, ( 5x - 7 ) and ( 3x + 17 ) are alternate interior angles, which means they are congruent, or equal:

5x - 7 = 3x + 17

Add 7 to each side:

5x = 3x + 24

Subtract 3x from both sides:

2x = 24

Divide each side by 2:

x = 12

Now we can see that ( 5x - 7 ) and ( 4y + 3 ) are supplementary angles, which means they will add up to 180 degrees:

5x - 7 + 4y + 3 = 180

Substitute 12 in for x to solve for y:

5 ( 12 ) - 7 + 4y + 3 = 180

60 - 7 + 4y + 3 = 180

combine like terms:

( 60 + ( -7 ) + 3 ) + 4y = 180

56 + 4y = 180

Subtract 56 from each side:

4y = 124

Divide each side by 4:

y = 31

6 0

3 years ago

3/4 ÷ 6/7 need help nowww

Lostsunrise [7]

Answer:

7/8 or in decimal form .87

Step-by-step explanation:

8 0

3 years ago

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, So

tatyana61 [14]

Answer:

The probability that at least 13 flights arrive late is 2.5196 $\times 10^{-6}$ .

Step-by-step explanation:

We are given that Southwest Air had the best rate with 80 % of its flights arriving on time.

A test is conducted by randomly selecting 18 Southwest flights and observing whether they arrive on time.

The above situation can be represented through binomial distribution;

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

where, n = number of trials (samples) taken = 18 Southwest flights

r = number of success = at least 13 flights arrive late

p = probability of success which in our question is probability that

flights arrive late, i.e. p = 1 - 0.80 = 20%

Let X = Number of flights that arrive late.

So, X ~ Binom(n = 18, p = 0.20)

Now, the probability that at least 13 flights arrive late is given by = P(X $\geq$ 13)

P(X $\geq$ 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)

= $\binom{18}{13}\times 0.20^{13} \times (1-0.20)^{18-13}+ \binom{18}{14}\times 0.20^{14} \times (1-0.20)^{18-14}+ \binom{18}{15}\times 0.20^{15} \times (1-0.20)^{18-15}+ \binom{18}{16}\times 0.20^{16} \times (1-0.20)^{18-16}+ \binom{18}{17}\times 0.20^{17} \times (1-0.20)^{18-17}+ \binom{18}{18}\times 0.20^{18} \times (1-0.20)^{18-18}$

= $\binom{18}{13}\times 0.20^{13} \times 0.80^{5}+ \binom{18}{14}\times 0.20^{14} \times 0.80^{4}+ \binom{18}{15}\times 0.20^{15} \times 0.80^{3}+ \binom{18}{16}\times 0.20^{16} \times 0.80^{2}+ \binom{18}{17}\times 0.20^{17} \times 0.80^{1}+ \binom{18}{18}\times 0.20^{18} \times 0.80^{0}$

= 2.5196 $\times 10^{-6}$ .

7 0

3 years ago

The length of a text messaging conversation is normally distributed with a mean of 2 minutes and a standard deviation of 0.5 min

zhenek [66]

Answer:

Let x be a random variable representing the length of a text messaging conversation, then

P(x > 3) = 1 - P(x < 3) = 1 - P(z < (3 - 2)/0.5) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275

5 0

3 years ago