Yong has 5 seed packets that each weigh 3/4 gram. How many grams of seeds does yong have in all

What is the relationship between 0.7 and 0.07

QveST [7]

Well there are a number of relationships, I'll name a few:

1) .7 > .07 (.7 is Greater than .07)
2) .07 < .7 (.07 is Less than .7)
3) .7 is ten times greater than .007

8 0

4 years ago

Read 2 more answers

Life f passes through points (1,14) and (10,10). Line g is perpendicular to f. What is the slope of line g

kolezko [41]

The slope of line g is 9/4.

Step-by-step explanation:

The line f passes through points (1,14) and (10,10).

Line g is perpendicular to f.

To find the slope of line f :

The formula for slope is given by,

$Slope = (y2-y2)/(x2-x1)$

Here, The point (1,14) is (x1.y1) and (10,10) is (x2,y2).

⇒ $(10-14) / (10-1)$

⇒ $-4 / 9$

Therefore, the slope of line f is -4/9.

To find the slope of line g :

Since, the line g is perpendicular to f, the slope of g will be the negative reciprocal of slope of line f.

⇒ slope of g = - (1 / slope of f).

⇒ $- (1 / (-4/9))$

⇒ $9/4$

∴ The slope of line g is 9/4.

7 0

3 years ago

Jonas’s vacuum cleaner uses 5/8 kilowatts of electricity per hour. His clothes dryer uses 4 11/16 of electricity per hour. How m

alekssr [168]

Hope this helps you out.

6 0

3 years ago

What is the sample space of the following spinner? A. Yellow green blue B. Red red yellow green blue blue C. 4 D. Red yellow gre

Vesnalui [34]

Answer:

S = { Red ,Yellow ,Blue, Red, Green,Blue}

n(S) = 6

Step-by-step explanation:

Explanation:

Sample space :

It is the total number of outcomes in the experiment.

Let the sample Space of the Spinner be 'S'

S = { Red ,Yellow ,Blue, Red, Green,Blue}

n(S) = 6

8 0

3 years ago

The lifespan of a lion in a particular zoo are normally distributed. The average lion lives 12.5 years the. Standard deviation i

Anika [276]

Answer:

0.1585

Step-by-step explanation:

Solution:-

The lifespan of a lion in a particular zoo is normally distributed with average lion lives:

Mean u = 12.5 years

Standard deviation s = 2.4 years

We are to use the empirical rule ( 68-95-99.7% ) to estimate the probability of a lion living between 5.3 and 10.1.

- The empirical rule ( 68-95-99.7% ) states:

P ( u - s < X < u + s ) = 68%

P ( u - 2s < X < u + 2s ) = 95%

P ( u - 3s < X < u + 3s ) = 99.7%

- The test have the following number of standard deviations (s):

u - s < X < u + s = 12.5 - 2.4 < X < 12.5 + 2.4 = 10.1 < X < 14.9

u - 2s < X < u + 2s = 12.5 - 4.8 < X < 12.5 + 4.8 = 7.7 < X < 17.3

u - 3s < X < u + 3s = 12.5 - 7.2 < X < 12.5 + 7.2 = 5.3 < X < 19.7

Hence,

P ( 10.1 < X < 14.9 ) = 0.68

P ( 7.7 < X < 17.3 ) = 0.95

P ( 5.3 < X < 19.7 ) = 0.997

- We need P ( X < 10.1 ) and P ( X < 5.3 ):

P ( X < 10.1 ) = [ 1 - P ( 10.1 < X < 14.9 ) ] / 2

= [ 1 - 0.68 ] / 2

= 0.16

P ( X < 5.3 ) = [ 1 - P ( 5.3 < X < 19.7 ) ] / 2

= [ 1 - 0.997 ] / 2

= 0.0015

Hence,

P ( 5.3 < X < 10.1 ) = P ( X < 10.1 ) - P ( X < 5.3 )

= 0.16 - 0.0015

= 0.1585

5 0

4 years ago