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Keith_Richards [23]

3 years ago

9

On monday, mary had 23 bananas. On tuesday, her family ate half of them. On wednesday they ate half of the remaining bananas. If

they continue each day, on which day of the week will there only be one banana left

1 answer:

telo118 [61]3 years ago

5 0

Answer:

Friday.

Step-by-step explanation:

The number of bananas is reducing from 23 at a rate of 50% of the previous day's number.

Given that on Monday, Mary had 23 bananas and we are asked that on which day there will be only 1 banana will left.

Let after d days the remaining bananas will become one.

Therefore, $23[1-\frac{50}{100} ]^{d} = 1$

⇒ $(0.5)^{d} =\frac{1}{23}$

Taking log both sides we get

$d \log 0.5 = \log (\frac{1}{23} )$

⇒ d = 4.523 days.

Therefore, on Friday the remaining number of bananas will become 1. (Answer)

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Find the product of<br>the sum of<br>3/5 and 1%<br>and

dybincka [34]

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Step-by-step explanation:

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3 years ago

An equation parallel and perpendicular to 4x+5y=19

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Answer:

Parallel line:

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Perpendicular line:

$y=\frac{5}{4}x-\frac{1}{2}$

Step-by-step explanation:

we are given equation 4x+5y=19

Firstly, we will solve for y

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$m=-\frac{4}{5}$

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we know that slope of two parallel lines are always same

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$y-1=-\frac{4}{5}(x-1)$

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Perpendicular line:

we know that slope of perpendicular line is -1/m

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Let's assume perpendicular line passes through (2,2)

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now, we can solve for y

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3 years ago

1.<br> Does the equation y = -x + 2 have a<br> positive or negative slope?

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This equation will have a Negative

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3 years ago

Read 2 more answers

A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th

umka2103 [35]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, <em>p</em> = 0.09.

An item is defective irrespective of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 9 and <em>p</em> = 0.09.

The repair cost of the item is given by:

$C=3X^{2}+X+2$

Compute the expected cost of repair as follows:

$E(C)=E(3X^{2}+X+2)$

$=3E(X^{2})+E(X)+2$

Compute the expected value of <em>X</em> as follows:

$E(X)=np$

$=4\times 0.09\\=0.36$

The expected value of <em>X</em> is 0.36.

Compute the variance of <em>X</em> as follows:

$V(X)=np(1-p)$

$=4\times 0.09\times 0.91\\=0.3276\\$

The variance of <em>X</em> is 0.3276.

The variance can also be computed using the formula:

$V(X)=E(Y^{2})-(E(Y))^{2}$

Then the formula of $E(Y^{2})$ is:

$E(Y^{2})=V(X)+(E(Y))^{2}$

Compute the value of $E(Y^{2})$ as follows:

$E(Y^{2})=V(X)+(E(Y))^{2}$

$=0.3276+(0.36)^{2}\\=0.4572$

The expected repair cost is:

$E(C)=3E(X^{2})+E(X)+2$

$=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73$

Thus, the expected repair cost is $3.73.

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3 years ago