Three things in any room that I can't measure the area of with square feet

Orlov [11]

Answer:

$1\ \text{nanometer}=4\times 10^8$

Step-by-step explanation:

Given : One nanometer equals about 0.00000004 inches.

To find : When writing this decimal as a single-digit integer multiplied by a power of ten, the single digit integer is and the power of ten ?

Solution :

$1\ \text{nanometer}=0.00000004$

$1\ \text{nanometer}=\frac{00000004}{100000000}$

$1\ \text{nanometer}=4\times 10^8$

Therefore, the decimal as a single-digit integer multiplied by a power of ten is $1\ \text{nanometer}=4\times 10^8$

4 0

3 years ago

Suzanne ran 334 miles in the morning and 412 miles in the afternoon. How many kilometers did she run altogether? Assume 1 mile

morpeh [17]

1193.6 because 334+412=746 and 746*1.6=1193.6

8 0

3 years ago

Which expression represents the greatest common factor (GCF) of 16 and 40?

zlopas [31]

the greatest common factor is 8

List the factors of 16 and 40:

16: 1,2,4,8,16

40: 1,2,4,5,8,10,20,40

8 is the greatest common factor

7 0

4 years ago

List fractions and decimals in order from least to greatest 2.3, 24\5,2.6

vaieri [72.5K]

I can answer this for you its 4.8,2.6,and 2.3

4 0

3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

3 years ago