What is the lateral area of a cylinder which has an element of 8 inches and a right section with a perimeter of 24 inches?

Thomas invested $8,500 for one year. Part of the money was invested at6% and the rest at 9%. The total interest earned was $667.

oee [108]

Okay so say that x=amount invested with 6% and y=amount invested with 9%
x+y=8,500 so > x=8500-y
6%=0.06 <span>9%=0.09</span>
0.06x +0.09y=667.5 (Substitute in x=8500-y so only numbers and y)
0.06(8500-y)+0.09y=667.5 (expand brackets)
510-0.06y+0.09y=667.5 (-510)
0.03y=117.5 (/0.03)
$3916.67=Y ->9%
X=8500-Y
x=$4583.33 ->6%

4 0

2 years ago

tom has 34 students in his class. 20 of them run around The track. 18 of them jump rope. How many do both?

Lera25 [3.4K]

Answer:

4

Step-by-step explanation:

20+18=38

38-34=4

hope this helped!!

3 0

2 years ago

Khalid and Jesse took different overnight trains. The graph shows the relationship between the total distance Khalid traveled an

Usimov [2.4K]

Answer:

Jesse - 150 miles per hour.

Step-by-step explanation:

On the graph, it shows that the rise is 140 and the run is 1. 140/1 is 140, making the slope for Khalid is 140. The equation (y = 150x) shows that the slope is 150 (y = mx). Since 150 is greater than (>) 140, Jesse's train is more faster than Khalid since she is moving at 150mph rather than 140mph.

Also, is this for Pearson Realize?

4 0

2 years ago

Find the x if the sequence 3, x, 4x/3 is (a)arithmetic and (b)geometric

jolli1 [7]

(a) When the sequence is arithmetic, sequential terms have a common difference.

... x - 3 = (4x/3) - x . . . . differences of sequential terms are equal

... (2/3)x = 3 . . . . . . . add 3-(1/3)x

... x = 9/2 . . . . . . . . . multiply by 3/2

(The arithmetic sequence is 3, 4.5, 6. The common difference is 3/2.)

(b) When the sequence is geometric, sequential terms have a common ratio.

... x/3 = (4x/3)/x . . . . . ratios of sequential terms are equal

... x^2 = 4x . . . . . multiply by 3x

... x = 4 . . . . . . . . divide by x. (the "solution" x=0 is extraneous)

(The geometric sequence is 3, 4, 16/3. The common ratio is 4/3.)

6 0

2 years ago

[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6

viktelen [127]

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

Height (in inches) of a 25 year old man is a normal random variable with mean $g=71$ and variance $o^{2} =6.25$ .

To find: (a) What percentage of 25 year old men are 6 feet, 2 inches tall

(b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X> $74in$ ]

P[X>74] = P[ $\frac{X-g}{o}$ > $\frac{74-71}{2.5}$ ]

= P[Z > 1.2]

= 1 - P[Z ≤ 1.2]

= 1 - Ф (1.2)

= 1 - 0.8849

= 0.1151

Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, $6ft$ $5in$ - X, $6ft$ ]

P[X > $6ft$ $5in$ I X > $6ft$ ] = P[X > 77 I X > 72]

= $\frac{P[X > 77]}{P[ X > 72]}$

= $\frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}] }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }$

= $\frac{P[Z >2.4]}{P[Z>0.4]}$

= $\frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }$

= $\frac{1-0.9918}{1-0.6554}$

= $\frac{0.0082}{0.3446}$

= 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

4 0

2 years ago