A train whose speed is 1.5 miles per minute travels y miles in x minutes. Express y in terms of x.

1. What is the decimal equivalent of 3/ 9? 0.2 0.3 0.4 0.5

viktelen [127]

0,3 is the decimal equivalent of 3/9

5 0

3 years ago

Can anyone help? It’s not 14, I’ll give Brainly + 15 pts

Julli [10]

Short side length = x

Medium side = x + 7

Long side = 5x

Total 49, so

x + x + 7 + 5x = 49

If you combine like terms:

7x + 7 = 49

Subtract 7 from both sides

7x = 42

Divide both sides by 7

x = 6 inches.

8 0

2 years ago

Let X be the temperature in at which a certain chemical reaction takes place, and let Y be the temperature in (so Y = 1.8X + 32)

Black_prince [1.1K]

Answer:

See explanation

Step-by-step explanation:

Solution:-

The random variable, Y be the temperature of chemical reaction in degree fahrenheit be a linear expression of a random variable X : The temperature in at which a certain chemical reaction takes place.

Y = 1.8*X + 32

- The median of the random variate "X" is given to be equal to "η". We can mathematically express it as:

P ( X ≤ η ) = 0.5

- Then the median of "Y" distribution can be expressed with the help of the relation given:

P ( Y ≤ 1.8*η + 32 )

- The left hand side of the inequality can be replaced by the linear relation:

P ( 1.8*X + 32 ≤ 1.8*η + 32 )

P ( 1.8*X ≤ 1.8*η ) ..... Cancel "1.8" on both sides.

P ( X ≤ η ) = 0.5 ...... Proven

Hence,

- Through conjecture we proved that: (1.8*η + 32) has to be the median of distribution "Y".

b)

- Recall that the definition of proportion (p) of distribution that lie within the 90th percentile. It can be mathematically expressed as the probability of random variate "X" at 90th percentile :

P ( X ≤ p_.9 ) = 0.9 ..... 90th percentile

- Now use the conjecture given as a linear expression random variate "Y",

P ( Y ≤ 1.8*p_0.9 + 32 ) = P ( 1.8*X + 32 ≤ 1.8*p_0.9 + 32 )

= P ( 1.8*X ≤ 1.8*p_0.9 )

= P ( X ≤ p_0.9 )

= 0.9

- So from conjecture we saw that the 90th percentile of "X" distribution is also the 90th percentile of "Y" distribution.

c)

- The more general relation between two random variate "Y" and "X" is given:

Y = aX + b

Where, a : is either a positive or negative constant.

- Denote, (np) as the 100th percentile of the X distribution, so the corresponding 100th percentile of the Y distribution would be : (a*np + b).

- When a is positive,

P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

= P ( a*X ≤ a*p_% )

= P ( X ≤ p_% )

= np_%

- When a is negative,

P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

= P ( a*X ≤ a*p_% )

= P ( X ≥ p_% )

= 1 - np_%

4 0

2 years ago

assume that when adults with smartphones are randomly selected 15 use them in meetings or classes if 15 adult smartphones are ra

Tanzania [10]

Answer:

The probability that at least 4 of them use their smartphones is 0.1773.

Step-by-step explanation:

We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.

Also, 15 adult smartphones are randomly selected.

Let X = Number of adults who use their smartphones

The above situation can be represented through the binomial distribution;

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

where, n = number of trials (samples) taken = 15 adult smartphones

r = number of success = at least 4

p = probability of success which in our question is the % of adults

who use them in meetings or classes, i.e. 15%.

So, X ~ Binom(n = 15, p = 0.15)

Now, the probability that at least 4 of them use their smartphones is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}$