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AURORKA [14]
3 years ago
5

How many real cube roots does –1 have?

Mathematics
1 answer:
Arlecino [84]3 years ago
7 0

Answer:

every number has a real root which is 1

this is not plagirism, the guy in the comments is copying and pasting so be careful. it can get you kicked out of your school.

Step-by-step explanation:

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Angelo bought apples and bananas at the fruit stand. He bought 20 pieces of fruits and spent $11.50. Apples cost $0.50 and banan
Cerrena [4.2K]
<h2>Steps</h2>

So for this system, one equation will represent the number of fruit and the other equation will represent the combined total of the fruit. (Let x = apples and y = bananas):

x+y=20\\0.5x+0.75y=11.50

Next, I will be using the substitution method to remove one of the variables. To do this, firstly subtract both sides by y in the first equation:

x=20-y\\0.5x+0.75y=11.50

Next, since we know that x is equal to 20 - y, substitute it into the second equation and solve for y:

0.5(20-y)+0.75y=11.50\\10-0.5y+0.75y=11.50\\10+0.25y=11.50\\0.25y=1.50\\y=6

Now that we have the value of y, substitute it into either equation to solve for x:

x+6=20\\x=14\\\\0.5x+0.75(6)=11.50\\0.5x+4.5=11.50\\0.5x=7\\x=14

<h2>Answer</h2>

<u>In short, Angelo bough 14 apples and 6 bananas.</u>

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The report "The New Food Fights: U.S. Public Divides Over Food Science" states that younger adults are more likely to see foods
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Solution :

$n_1=178, \hat p_1 = 0.48$

$n_2=427, \hat p_2 = 0.38$

$\hat p_1=\frac{x_1}{n_1}$

$x_1=n_1 \hat p_1$

   = 178 x 0.48

   = 85.44

    ≈ 85

$x_2=n_2 \hat p_2$

   = 427 x 0.38

   = 162.26

    ≈ 162

a). Yes, the sample sizes are large enough to use the large sample confidence interval so as to estimate the difference in the population proportions.

Let $\hat p_1 = 0.48, \ \ \hat p_2 = 0.38, n_1 = 178 \ \ n_2 = 427$

Where the $\text{subscript indicates}$ the age $18-29$ group and the 2 - subscript indicates the age $50-64$ group.

Since $n_1 \hat p_1 = 85.44$

$n_1(1- \hat p_1)=92.56, \ \ n_2\hat p_2 = 162.26, \ n_2(1-\hat p_2) = 264.74 $

are  all at least 10, the sample sizes are large enough to use the large sample confidence interval.

$P_0=\frac{x_1+x_2}{n_1+n_2}$

   $=\frac{85.44+162.26}{178+427}$

  = 0.409421

$P_0 = 0.4074$

$Q_0=1-P_0$

     = 1 - 0.4094

     = 0.5906

b). 90% confidence interval is

  $=\left( \hat p_1- \hat p_2 \pm \frac{z \alpha}{2} \times \sqrt{P_0Q_0\left(\frac{1}{n_1}+\frac{1}{n_2}}\right) \right)$

  $=\left( 0.48-0.38 \pm \frac{z \times 0.10}{2} \times \sqrt{0.4094 \times 0.5906\left(\frac{1}{178}+\frac{1}{427}}\right) \right)$

 $=(0.1 \pm z 0.05 \times 0.04387082)$

 $=(0.1 \pm 1.64 \times 0.0439)$

 $=(0.1 - 0.071996, 0.1+0.071996)$

 $=(0.028004, 0.171996)$

 $=(0.0280, 0.1720)$

c). Zero is not included in the confidence interval. Answer is no. The difference in the two population proportion are different from each other.

5 0
3 years ago
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