Multiplying Binomials: (−2h+9)(9h−2) = ?

michael borrowed $4,500 from the bank. the back charged a simple interest rate of 10.5% for the 3 years he had the back loan. wh

elixir [45]

Answer:

$1417.50+$4500= 5917.50

4 0

3 years ago

Matching quantities with units

Mrrafil [7]

Answer:

1. Volume = length x width x height => cubic centimetres.

2. Perimeter = 2(length) + 2(width) => metres.

3. Speed = Distance /time => metres per second.

4. Area = length x width => square metres.

Step-by-step explanation:

1. Volume = length x width x height

If the length, width and height are measured in centimetre (cm) , then, the unit of the volume will be:

Volume = cm x cm xcm = cm3 i.e cubic centimetres.

2. Perimeter = 2(length) + 2(width)

If the length and width are measured in meter (m), then, the unit of the perimeter will also be in metre (m).

3. Speed = Distance /time

If the unit of distance is measured in metre (m) and the time is measured in second (s), then , the unit of the speed will be:

Speed = Distance /time = m/s i.e metres per second.

4. Area = length x width

If the length and width are measured in metres (m), then, the unit of the area will be:

Area = m x m = m2 i.e square metres

8 0

3 years ago

Which statement about 307° is correct?

77julia77 [94]

Answer:

a hope help you stay happy and safe

4 0

3 years ago

Read 2 more answers

The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi

Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 51200, \sigma = 8200$

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{65000 - 51200}{8200}$

$Z = 1.68$

$Z = 1.68$ has a pvalue of 0.954

X = 55000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55000 - 51200}{8200}$

$Z = 0.46$

$Z = 0.46$ has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{48000 - 51200}{8200}$

$Z = -0.39$

$Z = -0.39$ has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{41000 - 51200}{8200}$

$Z = -1.24$

$Z = -1.24$ has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{61200 - 51200}{8200}$

$Z = 1.22$

$Z = 1.22$ has a pvalue of 0.889

X = 41200

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{41200 - 51200}{8200}$

$Z = -1.22$

$Z = -1.22$ has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0

2 years ago

A large university accepts 60% of the students who apply. Of the students the university accepts, 40% actually enroll. If 30,

kykrilka [37]

Answer:

7,200

Step-by-step explanation:

Let's start by calculating how many students the university accepts. The problem says they accept 60% of the students. 60% as a decimal is 0.6. To calculate 60% of something, you multiply .60 by whatever quantity. In this case, it would be the students.

Number of accepted students = $.6 * 30000$ = 18,000

The university accepts 18,000 students.

Now, the next problem is that of those accepted students, we need to find out how many actually enroll. The problem says they accept 40% of the accepted students. Applying the same principle,

Number of enrolled students = $.4 * 18000$ = 7,200

The number of students that actually enroll is 7,200.

Alternatively, $.6 * 30000 * .4$ would give you the same result.

Combining the pieces together, we are asking for 60% of 30,000. Then getting 40% of that 60%.

3 0

2 years ago