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zzz [600]
3 years ago
9

Jessica has just purchased a new purse in the shape of a rectangular box to match her favorite shoes. The purse has a length of

15 cm, a width of 6.4 cm, and a height of 8 cm. What is the volume of the purse?
Mathematics
1 answer:
Nostrana [21]3 years ago
6 0
The volume would be 768 cm^3
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8 0
3 years ago
Write a rational equation that can be solved by multiplying each side by 5(a+2)
qwelly [4]
What the answer would be is 10 :) hope that u are helped
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3 years ago
How many pages should the manufacturer advertise for each cartridge if it wants to be correct 99% of the time
PolarNik [594]

The manufacturer should advertise 13811 pages for each cartridge if it wants to be correct 99% of the time

z-score is the number of standard deviations from the mean value of the reference population

99%=0.99

The z-value associate with 0.99 is 2.33:

2.33 = (X - 12425) / 595:

X - 12425 = 2.33 * 595

X = (2.33 * 595) + 12425

X = 138111

Therefore, The manufacturer should advertise 13811 pages for each cartridge if it wants to be correct 99% of the time

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7 0
1 year ago
Solve the equation by factoring. 2x^2 + 15x + 25 = 0
Pepsi [2]

Answer:(2x+5)(x+5)=0

Step-by-step explanation:

2x^2+15x+25=0

Factorise we have

2x^2+10x+5x+25

(2x^2+10x)+(5x+25)=0

2x(x+5)+5(x+5)=0

(2x+5)(x+5)=0

7 0
3 years ago
Read 2 more answers
Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.
blsea [12.9K]

Answer:

a) Percentage of students scored below 300 is 1.79%.

b) Score puts someone in the 90th percentile is 638.

Step-by-step explanation:

Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

(a) If the average test score is 510 with a standard deviation of 100 points.

To find : What percentage of students scored below 300 ?

Solution :

Mean \mu=510,

Standard deviation \sigma=100

Sample mean x=300

Percentage of students scored below 300 is given by,

P(Z\leq \frac{x-\mu}{\sigma})\times 100

=P(Z\leq \frac{300-510}{100})\times 100

=P(Z\leq \frac{-210}{100})\times 100

=P(Z\leq-2.1)\times 100

=0.0179\times 100

=1.79\%

Percentage of students scored below 300 is 1.79%.

(b) What score puts someone in the 90th percentile?

90th percentile is such that,

P(x\leq t)=0.90

Now, P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90

P(Z< \frac{t-\mu}{\sigma})=0.90

\frac{t-\mu}{\sigma}=1.28

\frac{t-510}{100}=1.28

t-510=128

t=128+510

t=638

Score puts someone in the 90th percentile is 638.

5 0
3 years ago
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