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bearhunter [10]

2 years ago

8

A dressmaker needs to cut 6-inch pieces of ribbon from rolls of ribbon that are 9 feet in length. How many 6-inch pieces can the

dressmaker cut from 15 of these rolls of ribbon?

1 answer:

PSYCHO15rus [73]2 years ago

5 0

Answer:

270

Step-by-step explanation:

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A retailer sold an electric item to a customer at a loss of 10 percent the customer purchase it for rs 25425 including 13, perce

Mashutka [201]

Answer: Rs 25,000

Step-by-step explanation:

Given

Retailer sold an electric iron to a customer at a loss of 10%

Suppose the Cost price of iron is $x$

So, selling price is $0.9x$

This price must be equal to $25425$ + VAT

$\Rightarrow 0.9x=25425-0.13(0.9x)\\\Rightarrow 0.9x+0.117x=25425\\\Rightarrow 1.017x=25425\\\Rightarrow x=\text{Rs }25,000$

Thus, the cost price is $Rs.25,000$

5 0

2 years ago

Which formula can be used to describe the sequence?<br><br> -2 2/3,-5 1/3,-10 2/3, -21 1/3, -42 2/3

Kaylis [27]

Correct answer is third option ---C in edge

4 0

3 years ago

The average weight of a medium size egg is found to be 3.2 oz, with a standard deviation of 0.13 oz. What is the heaviest egg (t

Over [174]

Answer:

We are 95% confident that the average weight of the heaviest egg that is still medium size is 3.52 oz

Step-by-step explanation:

Mean = 3.2 oz, sd = 0.13 oz

Assuming there are 3 heavy eggs, n= 3, degree of freedom = n-1 = 3-1 = 2, t-value corresponding to 2 degrees of freedom and 95% confidence level is 4.303

Error margin = t×sd/√n = 4.303×0.13/√3 = 0.32 oz

Upper limit (heaviest egg that is still medium size) = mean + error margin = 3.2 + 0.32 = 3.52 oz

3 0

3 years ago

Correct answers only please!

Masteriza [31]

Answer:

Use the given formula :

I = P × R × t

I = 80 × 5/100 × 3

I = $ 12

5 0

2 years ago

The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =

gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy is given as :

$f_x(x) = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0$

Thus, the expected total claim amount $\mu$ = 1000

The variance of the total claim amount $\sigma ^2 = 1000^2$

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

$P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})$

$P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})$

$P(X>1100n) = P(Z> \dfrac{10*100}{1000})$

$P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345$

$\mathbf{P(X>1100n) = 0.158655}$

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

4 0

2 years ago