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Elena L [17]
3 years ago
8

There are bricks of silver and bricks of gold. When 5 silver bricks and 2 gold bricks are placed on a scale, the weight is 119 o

unces. When 4 silver bricks and 11 gold bricks are placed a scale the weight is 302 ounces. How much does each one weigh?
Mathematics
1 answer:
Illusion [34]3 years ago
4 0

Answer:

The weight of one gold brick is 15 ounces and the weight of one silver brick is 22 ounces.

Step-by-step explanation:

Given,

Weight of 5 silver bricks and 2 gold bricks = 119 ounces.

And Weight of 4 silver bricks and 11 gold bricks = 302 ounces.

We have to find out weight of each type of brick.

Solution,

Let the weight of 1 silver brick be x.

And the weight of 1 gold brick be y.

Since the weight of 5 silver bricks and 2 gold bricks is 119 ounces.

So we can frame it in equation as;

5x+2y=119\ \ \ \ \ equation\ 1

Again,

The weight of 4 silver bricks and 11 gold bricks is 302 ounces.

So we can frame it in equation as;

4x+11y=302\ \ \ \ \ equation\ 2

Now, multiplying equation 1 by 4 and equation 2 by 5 we get;

4(5x+2y)=119\times4\\\\20x+8y=476\ \ \ \ \ equation\ 3

5(4x+11y)=302\times5\\\\20x+55y=1510\ \ \ equation\ 4

Now subtract equation 3 from equation 4, we get;

(20x+55y)-(20x+8y)=1510-476\\\\20x+55y-20x-8y=1034\\\\47y=1034\\\\y=\frac{1034}{47}\\\\y=22

On substituting yhe value of y in equation 1, we get;

5x+2y=119\\\\5x+2\times22=119\\\\5x+44=119\\\\5x=119-44=75\\\\x=\frac{75}{5}\\\\x=15

Thus the weight of one gold brick is 15 ounces and the weight of one silver brick is 22 ounces.

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ivolga24 [154]

Answer:

(a) The mean and the standard deviation for the numbers of peas with green pods in the groups of 36 is 27 and 2.6 respectively.

(b) The significantly low values are those which are less than or equal to 21.8. And on the other hand, the significantly higher values are those which are greater than or equal to 32.2.

(c) The result of 15 peas with green pods is a result that is significantly​ low value.

Step-by-step explanation:

We are given that hybridization experiments are conducted with peas having the property that for​ offspring, there is a 0.75 probability that a pea has green pods.

Assume that the offspring peas are randomly selected in groups of 36.

The above situation can be represented as a binomial distribution;

where, n = sample of offspring peas = 36

            p = probability that a pea has green pods = 0.75

(a) The mean of the binomial distribution is given by the product of sample size (n) and the probability (p), that is;

                    Mean, \mu  =  n \times p

                                    =  36 \times 0.75 = 27 peas

So, the mean number of peas with green pods in the groups of 36 is 27.

Similarly, the standard deviation of the binomial distribution is given by the formula;

            Standard deviation, \sigma  =  \sqrt{n \times p \times (1-p)}

                                                  =  \sqrt{36 \times 0.75 \times (1-0.75)}

                                                  =  \sqrt{6.75}  =  2.6 peas

So, the standard deviation for the numbers of peas with green pods in the groups of 36 is 2.6.

             

(b) Now, the range rule of thumb states that the usual range of values lies within the 2 standard deviations of the mean, that means;

          \mu - 2 \sigma  =  27 - (2 \times 2.6)

                       =  27 - 5.2 = 21.8

          \mu + 2 \sigma  =  27 + (2 \times 2.6)

                       =  27 + 5.2 = 32.2

This means that the significantly low values are those which are less than or equal to 21.8.

And on the other hand, the significantly higher values are those which are greater than or equal to 32.2.

(c) The result of 15 peas with green pods is a result that is a significantly​ low value because the value of 15 is less than 21.8 which is represented as a significantly low value.

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