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2 years ago

Find the start time. Start time:____ P.M. Elapsed time: 2 1/4 hr End time: 4:44 P.M.

Mathematics

2 answers:

Artyom0805 [142]2 years ago

8 0

Answer:

2:30

Step-by-step explanation:

I think sorry if I'm wrong

iVinArrow [24]2 years ago

7 0

Answer:

2:29 PM

Step-by-step explanation:

You might be interested in

A new car is purchased for 17900 dollars. The value of the car depreciates at 12.25%

kiruha [24]

Answer:

Therefore the value of the car after 10 year will be = $4,845.34

Step-by-step explanation:

A new car is purchased for $17900. the value of the car depreciates at 12.25% per year.

Since the value of the car depreciates therefore the amount of the car is

$A= P(1-\frac{r}{100})^n}$

$=17900(1-\frac{12.25}{100})^{10}$ [ p = $17900, r = 12.25% and n =10 years]

= $ 4,845.34

Therefore the value of the car after 10 year will be = $4,845.34

3 0

2 years ago

Part F

mihalych1998 [28]

Answer:

The geometric mean of the measures of the line segments AD and DC is 60/13

Step-by-step explanation:

Geometric mean: BD² = AD×DC

BD = √(AD×DC)

hypotenuse/leg = leg/part

ΔADB: AC/12 = 12/AD

AC×AD = 12×12 = 144

AD = 144/AC

ΔBDC: AC/5 = 5/DC

AC×DC = 5×5 = 25

DC = 25/AC

BD = √[(144/AC)(25/AC)]

BD = (12×5)/AC

BD= 60/AC

Apply Pythagoras theorem in ΔABC

AC² = 12² + 5²

AC² = 144+ 25 = 169

AC = √169 = 13

BD = 60/13

The geometric mean of the measures of the line segments AD and DC is BD = 60/13

6 0

3 years ago

Nora drew a scale drawing of the middle school. The scale she use was 1 inch: 7 feet. The gym is 17 inches in the drawing. How i

emmainna [20.7K]

Answer: 119

Step-by-step explanation: if it’s 1:7,

then you’d have to do 17x7

5 0

2 years ago

The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep

Alchen [17]

Ooh, fun

geometric sequences can be represented as
$a_n=a(r)^{n-1}$
so the first 3 terms are
$a_1=a$
$a_2=a(r)$
$a_2=a(r)^2$

the sum is -7/10
$\frac{-7}{10}=a+ar+ar^2$
and their product is -1/125
$\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3$

from the 2nd equation we can take the cube root of both sides to get
$\frac{-1}{5}=ar$
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
$\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r$
subsituting -1/5 for ar
$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$
which simplifies to
$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for $ax^2+bx+c=0$
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$r=\frac{5 \pm \sqrt{25-16}}{4}$
$r=\frac{5 \pm \sqrt{9}}{4}$
$r=\frac{5 \pm 3}{4}$
so
$r=\frac{5+3}{4}=\frac{8}{4}=2$ or $r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$

use them to solve for the value of a
$\frac{-1}{5}=ar$
$\frac{-1}{5r}=a$
try for r=2 and 1/2
$a=\frac{-1}{10}$ or $a=\frac{-2}{5}$

test each
for a=-1/10 and r=2
a+ar+ar²= $\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$
it works

for a=-2/5 and r=1/2
a+ar+ar²= $\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$
it works

both have the same terms but one is simplified

the 3 numbers are $\frac{-2}{5}$ , $\frac{-1}{5}$ , and $\frac{-1}{10}$

6 0

3 years ago

Consider rolling two fair dice one 3-sided the other 5-sided

Ne4ueva [31]

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

$1\leq x\leq 3,\quad 1\leq y \leq 5$

We have $P(X=x)=\frac{1}{3}$ and $P(Y=y)=\frac{1}{5}$ , because the dice are fair.

Now we use the assumption of independence to claim that

$P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}$

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

2 in a unique way (1+1)
3 in two possible ways (1+2, 2+1)
4 in three possible ways
5 in three possible ways
6 in three possible ways
7 in two possible ways
8 in a unique way

This implies that the probabilities of the outcomes of $W=X+Y$ are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5

8 0

2 years ago