Which pair of terms can be used to represent any two consecutive odd numbers?

One gallon is equal to about 3.785 liters. What is this amount rounded to the nearest tenth? To the nearest hundredth? Show your

Leokris [45]

Answer:

3.8 ; 3.79 ; 3

Step-by-step explanation:

Given that:

1 gallon is equal to about 3.785 litres

3.785 to the nearest tenth :

Tenth digit = 7 ; round up 8 to 1 and add to 7

Hence,

3.785 = 3.8 (nearest tenth)

3.785 to the nearest hundredth :

Hundredth digit = 8 ; round up next digit 5 to 1 and add to 8

3.785 = 3.79 ( nearest hundredth)

What is the greatest number of whole liters of water you could pour into a one-gallon container without it overflowing?

The greatest Number of whole liters of water that could be poured into a 1 gallon container without it overflowing is 3 liters because, rounding up 3.785 to the nearest integer of 4 means we will exceed the maximum litres by about 0.215 gallons and hence, cause the container to overflow.

3 0

3 years ago

What is the solución to this equation X/5=30

Mama L [17]

Answer:

150

Step-by-step explanation:

x/5=30

x=30*5

x=150

Hoped this helped!

5 0

3 years ago

Read 2 more answers

Tia is investing $2500 that she would like to grow to $6000 in 10 years. At what annual interest rate, compounded quarterly, mus

timurjin [86]

To find the required interest rate, you need to solve the following equation for r:
$6000=2500(1+ \frac{r}{4} )^{40}$

7 0

3 years ago

Of all customers purchasing automatic garage-door openers, 75% purchase Swedish model. Let X = the number among the next 15 purc

Lelechka [254]

Answer:

a)

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

b)

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = 0.2737

d) μ = 15*0.75 = 11.25. σ² = 11.25*0.25 = 2.8125

Step-by-step explanation:

X is a binomial random variable with parameters n = 15, p = 0.75. Therefore

a)

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

b)

P(X>10) = P(X=11) + P(X=12)+ P(X=13)+P(X=14)+P(x=15)

$P(X=11) = {15 \choose 11} * 0.75^{11} * 0.25^4 = 0.2252$

$P(X=12) = {15 \choose 12} * 0.75^{12} * 0.25^3 = 0.2252$

$P(X=13) = {15 \choose 13} * 0.75^{13} * 0.25^2 = 0.1559$

$P(X=14) = {15 \choose 14} * 0.75^{14} * 0.25 = 0.0668$

$P(X=15) = {15 \choose 15} * 0.75^{15} = 0.0134$

Thus,

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X=9) + P(X=10)

$P(X=6) = {15 \choose 6} * 0.75^{6} * 0.25^9 = 0.0034$

$P(X=7) = {15 \choose 7} * 0.75^{7} * 0.25^8 = 0.0131$

$P(X=8) = {15 \choose 8} * 0.75^{8} * 0.25^7 = 0.0393$

$P(X=9) = {15 \choose 9} * 0.75^{9} * 0.25^6 = 0.0918$

$P(X=10) = {15 \choose 10} * 0.75^{10} * 0.25^{5} = 0.1652$

Thereofre,

$P(6 \leq X \leq 10) = 0.0034 + 0.0134 + 0.0393 + 0.0918 + 0.1652 = 0.2737$

d) μ = n*p = 15*0.75 = 11.25

σ² = np(1-p) = 11.25*0.25 = 2.8125

3 0

3 years ago

You purchase 26 "parking hours" that you can use over the next month to park

Nadusha1986 [10]

Answer:

The number of weekday hours are 5.

Step-by-step explanation:

Let the number of weekday hours be "x".

The total week hours are 26, then number of weekend hours is (26-x) .

The charges for weekdays are $2 per hour and for weekends $10 per hour.

The total money spent is $220.

$(2)(x) + (26-x)(10) = 220$

$8x = 40$

x = 5.

Thus weekday hours utilized are 5 hours, and weekend are (26-5) 21 hours.

4 0

3 years ago