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What is the area of a rectangle that is 3 1/2 x 6 1/2 ?

stiv31 [10]

3 1/2 * 6 1/2

3 1/2 to improper fraction = (3*2 + 1) / 2 = 7/2

6 1/2 to improper fraction = (6*2 + 1) / 2 = 13/2

3 1/2 * 6 1/2 = 7/2 * 13 /2 = 91 / 4

91 / 4

= 22 3 /4 square unit

8 0

4 years ago

Please read the attachment below, I need help on this one

Citrus2011 [14]

Answer:

A) 60% B)65%

Step-by-step explanation:

6 0

3 years ago

Matthew purchased a washer and dryer for $2,885 using an 18-month deferred payment plan with an interest rate of 26.12%. What is

-Dominant- [34]

For a deferred payment plan no interest is accrued for the deferment period.

If $85 is made each month, then in 18 months, 18 * 85 = $1,530 must have been paid off from the balance owed.

Thus, the remaining balance after the deferment period is $2,885 - $1,530 = $1,355.

3 0

3 years ago

Sailors once used a unit of length called a fathom to measure the depth of the water. Each fathom is equal to 6 feet. Which equa

ss7ja [257]

Answer:

y = x*(6 feet/1 fathom)

Step-by-step explanation:

We know the relation:

1 fathom = 6 feet.

We can write:

1 = (6 feet/1 fathom)

Now, remember that any number multiplied by 1 does not change.

Then suppose that we have a measure of 3 fathoms, then:

3 fathoms = (3 fathoms)*1

= (3 fathoms)*((6 feet/1 fathom)) = (3 fathoms/1 fathom)*(6feet)

= 3*(6 feet) = 18 feet.

So we changed the units by multiplying the original measure by (6 feet/1 fathom)

Then, if x is the depth of water in fathoms and y is the depth of the water in feet, we can write:

y = x*(6 feet/1 fathom)

3 0

3 years ago

The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide

kondor19780726 [428]

Answer:

99% confidence interval for the population standard deviation = (74.97 , 635.20).

Step-by-step explanation:

We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation

$\sigma$ = population standard deviation

n = sample of university professors = 5

Also, $s^{2}$ = $\frac{\sum (X-\bar X)^{2} }{n-1}$ = 144.5

So, 99% confidence interval for population standard deviation, $\sigma$ is;

P(0.2070 < $\chi^{2} __5_-_1$ < 14.86) = 0.99 {As the table of $\chi^{2}$ at 4 degree of freedom

gives critical values of 0.2070 & 14.86}

P(0.2070 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 14.86) = 0.99

P( $\frac{ 0.2070}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 14.86}{(n-1)s^{2} }$ ) = 0.99

P( $\frac{ (n-1)s^{2}}{14.86 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{0.2070 }$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{14.86 }$ , $\frac{ (n-1)s^{2}}{0.2070 }$ )

= ( $\frac{ (5-1) \times 144.5^{2}}{14.86 }$ , $\frac{ (5-1) \times 144.5^{2}}{0.2070 }$ )

= (5620.525 , 403483.092)

99% confidence interval for $\sigma$ = ( $\sqrt{5620.525}$ , $\sqrt{403483.092}$ )

= (74.97 , 635.20)

Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).

8 0

4 years ago