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3 years ago

5

Lulu tells her little brother, Jack, that she is holding 20 coins, all of which are either dimes or quarters. They have a value

of $4.10. She says she will give him the coins if he can tell her how many of each she is holding. Solve this problem for Jack.

1 answer:

Sunny_sXe [5.5K]3 years ago

6 0

Answer:

Lulu has 14 quarters and 6 dimes

Step-by-step explanation:

Let x be the number of dimes and y the number of quarters. The sum of quarters and dime is 20 coins. Then:

x + y = 20

We also know that the value of each dime is $ 0.10 and the value of each quarter is $0.25. The total value of the coins is $4.10. Then, the number of dimes times 0.10 plus the number of quarters times 0.25 will be 4.10:

0.10x + 0.25y = 4.10

So, we have a system of 2 equations and 2 unknowns:

x + y = 20

0.10x + 0.25y = 4.10

Let´s solve the first equation for y:

x + y = 20

subtract x from both sides

y = 20 - x

And now let´s replace y in the second equation:

0.10x + 0.25(20 - x) = 4.10

Apply distributive property

0.10x + 5 - 0.25x = 4.10

subtract 5 from both sides of the equation

-0.15x = -0.9

Divide by -0.15 both sides of the equation

x = 6

Now let´s calculate y:

y = 20 - x

y = 20 - 6 = 14

Lulu has 14 quarters and 6 dimes

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Read 2 more answers

The USDA conducted tests for salmonella in produce grown in California. In an independent sample of 252 cultures obtained from w

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Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 252$

The number that tested positive is $k_1 = 18$

The second sample size is $n_2 = 476$

The number that tested positive is $k_2 = 20$

The level of significance is $\alpha = 0.01$

Generally the first sample proportion is mathematically represented as

$\^ p _1 = \frac{k_1 }{ n_1 }$

=> $\^ p _1 = \frac{18 }{ 252 }$

=> $\^ p _1 = 0.071$

Generally the second sample proportion is mathematically represented as

$\^ p _2 = \frac{k_2 }{ n_2 }$

=> $\^ p _2 = \frac{20 }{ 476}$

=> $\^ p _2 = 0.042$

The null hypothesis is $H_o : p_1 - p_2 = 0$

The alternative hypothesis is $H_a : p_1 - p_2 \ne 0$

Generally the test statistics is mathematically represented

$z = \frac{ \^ p_1 - \^ p_2 - ( p_ 1 - p_2 )}{ \sqrt{\frac{\^ p_1 (1-\^ p_1)}{ n_1 } + \frac{\^ p_2 (1-\^ p_2)}{ n_2 } } }$

=> $z = \frac{ 0.071 - 0.042 - 0 }{ \sqrt{\frac{0.071 (1-0.071)}{ 252 } + \frac{0.042 (1-\^ 0.042)}{ 476 } } }$

=> $z = 1.56$

From the z table the area under the normal curve to the right corresponding to 1.56 is

$P(Z > 1.56 ) =0.05938$

Generally the p-value is mathematically represented as

$p-value = 2 * P(Z > 1.56 )$

=> $p-value = 2 * 0.05938$

=> $p-value = 0.1188$

From the value obtained we see that $p-value > \alpha$ hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

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2 years ago