All categories

3 years ago

Seventy three and sixty five hundredths

1 answer:

4 0

73.65

when and comes into play it means 73 is separated from the 65 hundredths

tens,ones AND 6 tenths and 5 hundredths

You might be interested in

Sara paid $15.95 to become a member at a gym. She then paid a monthly membership fee. Her total cost for 12 months was $735.95.

Step2247 [10]

735.95 - 15.95= 720, 720/12=60, monthly fee is 60

7 0

4 years ago

A plane flies 1225 miles with the wind in the same time it can fly 1100 miles into the wind. The speed of the plane in still air

iogann1982 [59]

Answer:

Step-by-step explanation:

v=d/t, t=d/v, since t=t

1225/(v+w)=1100/(v-w)

1225v-1225w=1100v+1100w

2325w=125v

w=125v/2325, given v=575

w=125(575)/2325

w=30.91 mph

8 0

3 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

We used Total - 1 because it is a probability without replacement

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

6 0

3 years ago

Jim bought 2 packs of batteries from a store. The price of each pack was the same. After he bought the batteries, his account ba

a_sh-v [17]

Answer:

Divide $23.94 by 2. 23.97/2=11.97. The change would have been -$11.97 :) hope this helps

Step-by-step explanation:

6 0

3 years ago

Let f(x) = (x − 1)2, g(x) = e−2x, and h(x) = 1 + ln(1 − 2x). (a) Find the linearizations of f, g, and h at a = 0. What do you no

sweet [91]

Answer:

Lf(x) = Lg(x) = Lh(x) = 1 - 2x

value of the functions and their derivative are the same at x = 0

Step-by-step explanation:

Given :

f(x) = (x − 1)^2,

g(x) = e^−2x ,

h(x) = 1 + ln(1 − 2x).

a) Determine Linearization of f, g and h at a = 0

L(x) = f (a) + f'(a) (x-a) ( linearization of f at a )

for f(x) = (x − 1)^2

f'(x ) = 2( x - 1 )

at x = 0

f' = -2

hence the Linearization at a = 0

Lf (x) = f(0) + f'(0) ( x - 0 )

Lf (x) = 1 -2 ( x - 0 ) = 1 - 2x

For g(x) = e^−2x

g'(x) = -2e^-2x

at x = 0

g(0) = 1

g'(0) = -2e^0 = -2

hence linearization at a = 0

Lg(x) = g ( 0 ) + g' (0) (x - 0 )

Lg(x) = 1 - 2x

For h(x) = 1 + ln(1 − 2x).

h'(x) = -2 / ( 1 - 2x )

at x = 0

h(0) = 1

h'(0) = -2

hence linearization at a = 0

Lh(x) = h(0) + h'(0) (x-0)

= 1 - 2x

Observation and reason

The Linearization is the same in every function i.e. Lf(x) = Lg(x) = Lh(x) this is because the value of the functions and their derivative are the same at x = 0

8 0

3 years ago