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soldi70 [24.7K]
3 years ago
14

A shopkeeper with a basket of eggs finds that if he sells 3 eggs at a time there is only one egg left if he sells four eggs at a

time there is again 1 egg left however if the trader sells 7 is at a time there is no egg left is the capacity of the basket is 108 then find how many days are there in the basket explain with reasoning.
​
Mathematics
2 answers:
Agata [3.3K]3 years ago
7 0

Answer:

Step-by-step explanation:

add a picture of the question please

erica [24]3 years ago
4 0

Solution:- A shopkeeper with a basket of eggs finds that if he sells 3 eggs at a time there is only one egg left if he sells four eggs at a time there is again 1 egg left however if the trader sells 7 is at a time there is no egg left is the capacity of the basket is 108

LCM of 3 and 4 = 3×4 = 12

The capacity of the basket is 100.

So, there may be:-

= 12 × 8 +1

= 96 + 1

= 97 eggs.

But, 97 is not divisible by 7.

Again, there may be:

12 × 7 +1 = 85 eggs.

12 × 6 +1 = 73 eggs.

12 × 5 +1 = 61 eggs.

12 × 4 +1 = 49 eggs.

Among these numbers 85 , 73 and 61 are not divisible by 7. But 49 is divisible by 7.

So, the number of eggs in the basket is 49.

[Answer = 49]

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What is the solution to the equation log(2x + 4) = 2? Round to the nearest thousandth, if necessary.
Kazeer [188]

Answer: 48

Step-by-step explanation:

log_{b}M=n--> M=b^{n} \\log(2x+4)=2\\b=10, M=2x+4, n= 2 \\\\2x+4=10^{2} \\2x+4=100\\2x=96\\x=48

5 0
3 years ago
The breaking strength of a cable is assumed to be lognormally distributed with a mean of 75 and standard deviation of 20. • If t
son4ous [18]

Answer:

If the load of magnitude 50 is hung from the cable, determine the probability of failure of the cable?

There is a 10.56% probability of failure of the cable.

If the load can take values 30, 50, and 70 with probabilities 0.2, 0.35, and 0.45 respectively, determine the probability of failure of the cable?

There is a 15.87% probability of failure of the cable.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The breaking strength of a cable is assumed to be lognormally distributed with a mean of 75 and standard deviation of 20. This means that \mu = 75, \sigma = 20.

If the load of magnitude 50 is hung from the cable, determine the probability of failure of the cable?

This is the pvalue of Z when X = 50.

Z = \frac{X - \mu}{\sigma}

Z = \frac{50 - 75}{20}

Z = -1.25

Z = -1.25 has a pvalue of 0.1056.

There is a 10.56% probability of failure of the cable.

If the load can take values 30, 50, and 70 with probabilities 0.2, 0.35, and 0.45 respectively, determine the probability of failure of the cable?

Now we have that

X = 0.2(30) + 0.35(50) + 0.45(70) = 55

Z = \frac{X - \mu}{\sigma}

Z = \frac{55 - 75}{20}

Z = -1

Z = -1 has a pvalue of 0.1587.

There is a 15.87% probability of failure of the cable.

3 0
4 years ago
(C) A bakery sells 9 types of sandwiches. Here are their calorie amounts: 562.568. 590. 599. 611.621. 623. 641, 938. Which measu
Ludmilka [50]

Answer:

611

Step-by-step explanation:

The mean is 632.9 which is a lot higher than 611, and is actually higher than the penultimate value 632, and if presented as a general single value, it may mislead customers into thinking that the sandwiches have a high calorific value. We don’t know how many sandwiches of different types there are, and there may only be a few with high values. So I would use the median 611 as a typical value for the number of calories.

6 0
3 years ago
Her senior year of college, Giselle decides to live off campus. In order to cover rent, she wants to get a job as a waitress.
SCORPION-xisa [38]
She would have to work at least 12 hours.

At Chili’s, she would be paid 11.75x + 33 per week, with x being the number of hours she worked.

At the Cheesecake Factory, Giselle would be paid 14.50x per week, with x being how many hours she worked.

We want to know how many hours Gisele would have to work for her pay at the Cheesecake Factory to be more than her pay at Chili’s

The inequality 11.75x + 33 < 14.50x is what has to be set up to solve the problem. At how many hours will the pay on the left be less than the pay on the right?

11.75x + 33 < 14.50x
33 < 2.75x
12 < x

So Giselle has to work greater 12 hours a week for her to make more money at the Cheesecake Factory.
5 0
3 years ago
Hannah wants to buy a new shirt the original price is 32.75 but it is on sale for 40% off . what is sale price of the shirt
kati45 [8]
1) 32.75 times .40= 13.10
2) 32.75-13.10=19.65
The sale price is $19.65
4 0
4 years ago
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