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OleMash [197]
3 years ago
13

Which mathematical property is demonstrated?

Mathematics
2 answers:
Kazeer [188]3 years ago
7 0
40 * (25 * 74) = (40 * 25) * 74

40 * (1850) = (1000) * 74

40 * (1850) = (1000) * 74

74,000 = 74,000

Knowing this, we can easily conclude that this is demonstrating the associative property of multiplication. The associative property of multiplication states that the product is the same regardless of their grouping. In this case, after simplifying the expression given, we can see that indeed the products are the same hence that fact that they have different grouping. 
ExtremeBDS [4]3 years ago
5 0

Answer:

Step-by-step explanation:

The associative property of multiplication states that the product is the same regardless of their grouping. In this case, after simplifying the expression given, we can see that indeed the products are the same hence that fact that they have different grouping

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Find the compound Interest for Rs.18,800 ,calculated for 2 years at 13% rate of Interest compounded annually
Vinvika [58]

For compound interest, the formula is given below:

Amount = P(1+\frac{r}{100} )^{n}

Here, P = 18,800

n = 2

r = 13/100

So, Amount = 18,800(1+\frac{13}{100} )^{2}

18,800(1.13)^{2}

= 18,800 × 1.2769

= 24005.72

Compound Interest = Amount - Principal

Compound Interest = 24005.72 - 18800

= 5205.72

Hence, the compound interest for Rs.18,800, calculated for 2 years at 13% rate of interest compounded annually is Rs.5205.72.


4 0
3 years ago
Find the area of a triangle with a base length of 6 units and a height of 7 units. PLEASE SHOW YOUR WORK! (10 points)
NARA [144]

Answer:

21 units

Step-by-step explanation:

The formula for solving a triangle is (b×h)/2

So that means 7 × 6 = 42 so then you want to divide 42 by 2

42÷2= 21

8 0
3 years ago
Read 2 more answers
The probability is 0.36 that the amount spent on a randomly selected child will be between what two value equidistant from the m
erastova [34]

Assuming a standard normal distribution, the positive value of interest will be that corresponding to the 50 + 36/2 = 68th percentile. A suitable probability calculator will give that value as 0.46770.


The values that are within ±0.46770 standard deviations from the mean will have a probability of 0.36.

4 0
4 years ago
The normal distribution An automobile battery manufacturer offers a 31/54 warranty on its batteries. The first number in the war
seraphim [82]

Answer:

1) if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2) a standard deviation of 6.0843 results in a 1.07% replacement rate

3) using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

Step-by-step explanation:

based on the given data;

x will represent the random variable such that the lifetime of its auto batteries, is normally distributed with a mean of 45 months and a standard deviation of 5.6 months

so

x → N( U = 45, ∝ = 5.6)

Under the warranty, if a battery fails within 31 months of purchase, the manufacturer replaces the battery at no charges to the consumer.

if the battery fails after 31 months but within 54 months, the manufacturer provides a prostrated credit towards the purchase of anew battery

1) If the manufacturer's assumptions are correct,

p(x < 3) = p( [x-u / ∝ ] < [ 31-45 / 5.6] )

= p( z < -2.5 )

using the standard normal table,

value of z = 0.0062 ≈ 0.62%

so if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2)

The company finds that it is replacing 1.07% of its batteries free of charge. It suspects that its assumption standard deviation of the life of its batteries is incorrect, so a standard deviation of ? results in a 1.07%

so lets say;

p ( x < 31 ) = ( 1.07%) = 0.0107

p ( [x-u / ∝ ] < [ 31-45 / ∝] ) = 0.0107

now from the standard table

-2.301 is 1.07%

so

( 31 - 45 / ∝ ) = -2.301

-14 / ∝ = -2.301

∝ = -14 / - 2.301

∝ = 6.0843

therefore a standard deviation of 6.0843 results in a 1.07% replacement rate

3)

Using the revised standard deviation for battery life, what percentage of the manufacturer's batteries don't free replacement but do qualify for the prorated credit?

p( 31 < x < 54 ) = p ( [31 - u / ∝ ] < [ x-u / ∝]  < [ 54 - 45 / ∝] )

= p ( [31 - 45 / 6.0843 ] < [ x-u / ∝]  < [ 54 - 45 / 6.0843] )

= p ( -2.301 < z < 1.4792 )

= p(Z < 1.5) - p(Z < -2.3)

= 0.9393 - 0.0108

= 0.919 ≈ 91.9%

therefore using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

8 0
3 years ago
Please help me with this question please!!!
rodikova [14]
Answer is 182. Just substitute the x and y value into the equation.
3 0
3 years ago
Read 2 more answers
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