Answer:
The total money Mr. Sharma originally had was Rs. 1, 95, 000
Step-by-step explanation:
Let the original money Mr. Sharma had = (K) Rs
So, the share of his son = K/3
Share of his daughter = K/5
Share of his wife = Rs. 91,000
Now, according to the question:
Share of Son + Share of Daughter + Share of Wife = Mr. Sharma's total money
or,
or,
or,
⇒
⇒
or, K = 1,95,000
Hence, the total money Mr. Sharma originally had was Rs. 1, 95, 000
Answer:
Step-by-step explanation:
We have been given that Wren’s first display box is 6 inches long, 9 inches wide, and 4 inches high. We are asked to find the volume of display box.
We know that the box is in form of cuboid, so its volume would be length times width times height.
Therefore, the volume of the display box is 216 cubic inches.
Answer:
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.
Step-by-step explanation:
We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.
Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.
Let X = <u><em>back-to-college family spending on electronics</em></u>
SO, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean family spending = $237
= standard deviation = $54
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)
P(X < $150) = P( <
) = P(Z < -1.61) = 1 - P(Z
1.61)
= 1 - 0.9463 = <u>0.0537</u>
The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)
P(X > $390) = P( >
) = P(Z > 2.83) = 1 - P(Z
2.83)
= 1 - 0.9977 = <u>0.0023</u>
The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)
P($120 < X < $175) = P(X < $175) - P(X $120)
P(X < $175) = P( <
) = P(Z < -1.15) = 1 - P(Z
1.15)
= 1 - 0.8749 = 0.1251
P(X < $120) = P( <
) = P(Z < -2.17) = 1 - P(Z
2.17)
= 1 - 0.9850 = 0.015
The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.
Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>