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3 years ago

14

Bishwant and sunayana cimplete a oiece of a work in 6 days working together.if bishwant alime can complete it n 10 days.in how m

any days would sunayana alon3 complete the work

1 answer:

alina1380 [7]3 years ago

4 0

Step-by-step explanation:

sunaina's alona work=

1/6-1/10

10-6/60

4/60

1/15

therefore sunaina can alone do the work in 15 days

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If a cube with the length of the side of 4 cm is cut into smaller cubes with the length of the side of 1 cm, then what is the pe

ValentinkaMS [17]

Answer:

300 %

Step-by-step explanation:

Length of the larger cube =4 cm

So volume $V=side^3=4^3=64cm^3$

Length of the smaller cube = 1 cm

Volume of the smaller cube $V=side^3=1^3=1cm^3$

So total number of smaller cube $=\frac{64}{1}=64$

Surface area of the larger cube $A=6\times side^2=6\times 4^2=144cm^2$

Surface area of the 64 smaller cube $A=64\times 6\times side^2=64\times 6\times 1^2=384cm^2$

So percentage increase in surface area $=\frac{384-96}{96}\times 100=300$ %

6 0

3 years ago

Sharon wants to buy a shirt that costs $50. the sales tax is 5%. how much is the sales tax? what is her total cost for the shirt

maria [59]

5% of $50 = 5/100 x 50 = $2.50
Total = 50 + 2.50 = $52.50

The saes tax is $2.50; her total cost of the shirt is $52.50

6 0

3 years ago

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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago

Ken said that he is going to reduce the number of calories that he eats during the day. Ken's trainer asked him to start off sma

abruzzese [7]

2.200 or 2,200????????????

4 0

3 years ago

What is 6+2-8? make sure u give my my asnwer

yanalaym [24]

Answer: 0

Step-by-step explanation: 8-8 equals... ZERO GOOD JOB GUYS!

7 0

3 years ago

Read 2 more answers