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KengaRu [80]
3 years ago
8

In a village the number of houses and the number of flats are in the ratio 9:5 the number of flats and the number of bungalows T

here are 30 bungalows in a village How many houses are there in a village
Mathematics
1 answer:
Makovka662 [10]3 years ago
5 0

Answer:

Total no. of houses = 140

Step-by-step explanation:

Let there are 9x flats and 5x bungalows.

Total bungalows = 50

It means,

5x = 50

x = 10

No of flats = 9x

= 9(10)

= 90

Total houses = no. of flats + no. of bungalows

= 50+90

= 140

So, there are 140 houses in the village.

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The first day, Brian saw 31 birds and twice as many squirrels as birds. The second day, Brian saw 20 birds and 42 squirrels. Whi
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Answer:  A good estimation would be about 150.

Step-by-step explanation:

To estimate we can round instead of using exact numbers.

First day he saw about 30 birds and twice as many squirrels, meaning about 60 squirrels.  This is about 90 total when added together.

Second day he saw 20 birds and about 40 squirrels so about 60 when added together.

Add day one and day two totals for...

90+60=150

8 0
3 years ago
Some people think it is unlucky if the 13th day of month falls on a Friday. show that in that there year (non-leap or leap) ther
Vlad1618 [11]
<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths. Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday. Jan - Friday Feb - Monday Mar - Monday Apr - Thursday May - Saturday Jun - Tuesday Jul - Thursday Aug - Sunday Sep - Wednesday Oct - Friday Nov - Monday Dec - Wednesday Now let's count how many times for each weekday, the 13th falls there. Sunday - 1 Monday - 3 Tuesday - 1 Wednesday - 2 Thursday - 2 Friday - 2 Saturday - 1 The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc. So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get: Jan - Friday Feb - Monday Mar - Tuesday Apr - Friday May - Sunday Jun - Wednesday Jul - Friday Aug - Monday Sep - Thursday Oct - Saturday Nov - Tuesday Dec - Thursday And the weekday totals are: Sunday - 1 Monday - 2 Tuesday - 2 Wednesday - 1 Thursday - 2 Friday - 3 Saturday - 1 And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year. And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
5 0
3 years ago
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8 0
3 years ago
Find the missing dimensions: Use the scale factor 1/4:3
katrin [286]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
A private and a public university are located in the same city. For the private university, 1038 alumni were surveyed and 647 sa
Snezhnost [94]

Answer:

The difference in the sample proportions is not statistically significant at 0.05 significance level.

Step-by-step explanation:

Significance level is missing, it is  α=0.05

Let p(public) be the proportion of alumni of the public university who attended at least one class reunion  

p(private) be the proportion of alumni of the private university who attended at least one class reunion  

Hypotheses are:

H_{0}: p(public) = p(private)

H_{a}: p(public) ≠ p(private)

The formula for the test statistic is given as:

z=\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}} where

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  • p is the pool proportion of p1 and p2 (\frac{808+647}{1311+1038}=0.619)
  • n1 is the sample size of the alumni from public university (1311)
  • n2 is the sample size of the students from private university (1038)

Then z=\frac{0.616-0.623}{\sqrt{{0.619*0.381*(\frac{1}{1311} +\frac{1}{1038}) }}} =-0.207

Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.  

6 0
3 years ago
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