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den301095
21 day ago
11

Solve the following multiplication and division problems. a. 8 T. 1,398 lb. 14 oz. × 6 b. 349 lb. 6 oz. ÷ 130 c. 6 T. 294 lb. ÷

3,071
Mathematics
2 answers:
AnnZ [12.3K]21 day ago
6 0
a) 278,382 oz b) 43 oz c) 64.1 oz
Inessa [12.5K]21 day ago
5 0
A. 278,382 ounces B. 43 ounces C. 64.1 ounces
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The volume of the rectangular prism is 32 cubic feet more than the volume of the right triangular prism. Find the volume of each
tester [12383]

Answer:

Step-by-step explanation:

Denote the rectangular prism's volume as VR and the volume of the right triangular prism as VT. Given that the rectangular prism's volume exceeds that of the triangular prism by 32 cubic feet, we establish that VR = 32 + VT.

Calculating VR: VR = length * width * height = 6*x*3.

This simplifies to VR = 18x ft³.

For the triangular prism, VT can be calculated as Length * width * Height / 2, yielding VT = (7 * x * 4) / 2.

Thus, VT = 28x / 2, which simplifies to VT = 14x ft³.

Equating VR to 32 + VT allows the substitution: 18x = 32 + (14x).

Now, simplify the equation:

Combine like terms: 18x - 14x = 32.

This results in 4x = 32.

Dividing both sides by 4:

x = 8.

The volume for the rectangular prism comes to 18x, which equals 18*8.

The volume of the rectangular prism calculates to 144 ft³.

The volume for the right triangular prism calculates to 14x, which equals 14*8.

The volume of the right triangular prism calculates to 112 ft³.

4 0
1 month ago
Read 2 more answers
The probability of drawing two aces from a standard deck is 0.0059. We know this probability, but we don't know if the first car
Inessa [12570]

Answer:Option C is correct.

C. The events are independent because the first ace was replaced before drawing the second ace.

Step-by-step explanation:

It is given that the probability of drawing two aces from a standard deck is 0.0059

If the first card is drawn and replaced, it alters the probability. Performing draws with replacement makes each event independent of one another

The chance of getting an ace on the first draw is 4/52, and with the replacement, the probability remains 4/52 for the second draw

Therefore, if we consider the first and second draws as event A and event B respectively, these events are independent

C. The events are independent because the first ace was replaced before drawing the second ace.

4 0
11 days ago
Aubrey's dinner cost \$85$85dollar sign, 85. She tips the waitstaff 30\%30%30, percent for excellent service.
tester [12383]
Aubrey's dinner totaled $85. For excellent service, she tips the waitstaff 30%. The tip amount is $25.5. To calculate the tip, we start with the total cost of the dinner, which is $85, and apply the 30% tip rate. So, 30% of $85 is calculated as: Tip = 30% of 85. This results in Aubrey tipping the waitstaff $25.5.
4 0
1 month ago
Wes and Libbie each surveyed groups of students to find whether they would prefer going to an art museum or to the zoo. Wes foun
lawyer [12517]

Answer:

A.

Step-by-step explanation:

7 0
2 months ago
Let X denote the data transfer time (ms) in a grid computing system (the time required for data transfer) between a "worker" com
Inessa [12570]

Answer:

a. Alpha equals 3.014 while beta equals 12.442

b. The likelihood that the data transfer duration surpasses 50ms is 0.238

c. The chance that data transfer time falls between 50 and 75 ms is 0.176

Step-by-step explanation:

a. Given the data, the mean and standard deviation for the random variable X are 37.5 ms and 21.6, respectively.

Thus, E(X)=37.5 and V(X)=(21.6)∧2  

To find alpha, we need to apply the formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To determine beta, the following formula is employed:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. With E(X)=37.5 and V(X)=(21.6)∧2,  

Hence, P(X>50)=1−P(X≤50)

To find the probability of data transfer time exceeding 50ms, we use the formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The chance of data transfer time exceeding 50ms is 0.238

c. With E(X)=37.5 and V(X)=(21.6)∧2,  

Thus, P(50<X<75)=P(X<75)−P(X<50)  

To find the probability that data transfer time is between 50 and 75 ms, we apply the formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time falls between 50 and 75 ms is 0.176

6 0
1 month ago
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