A typical music CD has a diameter of 5 inches. How does the diameter compare to the circumference of the CD?

If there are three leaves of lettuce on hectors sandwich, what is the cost of lettuce on his sandwich?Explain

ANTONII [103]

Answer:

0.06$

Step-by-step explanation:

7 0

2 years ago

HELP PLZ NEED HELP SHOW WORK TOO

monitta

15. 3x - 2y = -2
3x - 3x - 2y = -3x - 2
   -2y = -3x - 2
      -2    -2
   y = 1.5x + 1

y - y₁ = m(x - x₁)
y - 3 = ⁻²/₃(x - (-2))
y - 3 = ⁻²/₃(x + 2)
y - 3 = ²/₃(x) - ²/₃(2)
y - 3 = ⁻²/₃x - 1¹/₃
   + 3 + 3
   y = ⁻²/₃x + 1²/₃

16. 230 = 0.2s + 150
    - 150    - 150
      80 = 0.2s
      0.2    0.2
    400 = s

17. y = 2x + 2

5 0

3 years ago

Hi, can someone please answer this question? Thanks!!!

cricket20 [7]

Answer:

sorry im not so sure abut the answer

Step-by-step explanation:

.....

8 0

3 years ago

Solve 1.25x + 1.50y = 28.25.

fiasKO [112]

To find the x-intercept, substitute in

0

for

y

and solve for

x

. To find the y-intercept, substitute in

0

for

x

and solve for

y

.

x-intercept(s):

(

22.6

,

0

)

(

22.6

,

0

)

y-intercept(s):

(

0

,

18.8

¯

3

)

4 0

2 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

2 years ago