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4 years ago

What is equivalent to (1/4) to the -3 power

Mathematics

2 answers:

ale4655 [162]4 years ago

5 0

Exact Form: -3/4 (fraction)

Decimal Form: -0.75

Lyrx [107]4 years ago

3 0

Exact Form: -3/4 (fraction)

Decimal Form: -0.75

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Of a group of randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wa

Wittaler [7]

Answer:

We are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

Step-by-step explanation:

We are given that in a group of randomly selected adults, 160 identified themselves as executives.

n = 160

Also we are given that 42 of executives preferred trucks.

So the proportion of executives who prefer trucks is given by

p = 42/160

p = 0.2625

We are asked to find the 95% confidence interval for the percent of executives who prefer trucks.

We can use normal distribution for this problem if the following conditions are satisfied.

n×p ≥ 10

160×0.2625 ≥ 10

42 ≥ 10 (satisfied)

n×(1 - p) ≥ 10

160×(1 - 0.2625) ≥ 10

118 ≥ 10 (satisfied)

The required confidence interval is given by

$p \pm z\times \sqrt{\frac{p(1-p)}{n} }$

Where p is the proportion of executives who prefer trucks, n is the number of executives and z is the z-score corresponding to the confidence level of 95%.

Form the z-table, the z-score corresponding to the confidence level of 95% is 1.96

$p \pm z\times \sqrt{\frac{p(1-p)}{n} }$

$0.2625 \pm 1.96\times \sqrt{\frac{0.2625(1-0.2625)}{160} }$

$0.2625 \pm 1.96\times 0.03478$

$0.2625 \pm 0.06816$

$0.2625 - 0.06816, \: 0.2625 + 0.06816$

$(0.1943, \: 0.3306)$

$(19.43\%, \: 33.06\%)$

Therefore, we are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

5 0

4 years ago

An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number

vovikov84 [41]

Answer:

The number of key rings sold on a particular day when the total profit is $5,000 is 4,000 rings.

Step-by-step explanation:

The question is incomplete.

An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number of key rings is given by

$P=n^2-2n-3$

where n is the number of key rings in thousands.

Find the number of key rings sold on a particular day when the total profit is $5,000.

We have the profit defined by a quadratic function.

We have to calculate n, for which the profit is $5,000.

$P=n^2-2n-3=5\\\\n^2-2n-8=0$

We have to calculate the roots of the polynomial we use the quadratic equation:

$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\n= \frac{-2\pm\sqrt{4-4*1*(-8)}}{2}= \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{36}}{2} =\frac{-2\pm6}{2} \\\\n_1=(-2-6)/2=-8/2=-4\\\\n_2=(-2+6)/2=4/2=2$