Predict how much money can be saved without having a negative actual net income.

Mathematics

1 answer:

satela [25.4K]3 years ago

6 0

$170 can be saved resulting in an actual net income of $0.

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Can someone help me with this math question? First to answer will get Brainliest! :)

svetlana [45]

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2 years ago

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Which expression is equivalent to...

torisob [31]

Answer:

$x^{\frac{2}{7} } y^{-\frac{3}{5} } \\$ i.e answer A.

Step-by-step explanation:

This question involves knowing the following power/exponent rule:

$\sqrt[n]{x^m} = x^\frac{m}{n} \\\\so \sqrt[7]{x^2} = x^\frac{2}{7} \\\\and \\\\ \sqrt[5]{y^3} = y^\frac{3}{5} \\$

Next, when a power is on the bottom of a fraction, if we want to move it to the top, this makes the power become negative.

so the y-term, when moved to the top of the fraction, becomes:

$y^{-\frac{3}{5} } \\$

So the answer is: $x^{\frac{2}{7} } y^{-\frac{3}{5} } \\$

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2 years ago

The proportion of households in a region that do some or all of their banking on the Internet is 0.31. In a random sample of 100

Alenkasestr [34]

Answer:

Approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is less than 0.0005% .

Step-by-step explanation:

We are given that let X be the number that do some or all of their banking on the Internet.

Also; Mean, $\mu$ = 310/1000 or 0.31 and Standard deviation, $\sigma$ = 14.63/1000 = 0.01463 .

We know that Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

Probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is given by P(X <= 130/1000);

P(X <=0.13) = P( $\frac{X-\mu}{\sigma}$ <= $\frac{0.13-0.31}{0.01463}$ ) = P(Z <= -12.303) = P(Z > 12.303)