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Lubov Fominskaja [6]
2 years ago
11

In a recent survey, 15 skydivers said that they have never had a skydiving accident. This was 60% of the total number of skydive

rs that were surveyed. What was the total number of skydivers surveyed?
Mathematics
1 answer:
aleksandrvk [35]2 years ago
5 0

Answer:

total sky divers is 25

Step-by-step explanation:

60% = 15 skydivers

divide by 6 both sides

10% = 2.5

times by 4 both sides

40%=10

_________

60%+40%=100%

15+10=25

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Answer:

It represents a linear function and there will be 216 ants after 1 year

8 0
2 years ago
Which expression is equivalent to 3y+4+y
Colt1911 [192]
The answer to this question is 4y+4
8 0
3 years ago
Adult tickets to a play cost $22 tickets for kids cost $15. Tickets for a group of 11 cost a total of $228. How many children an
Zina [86]

Answer:Number of children = 2 and number of adults = 9 attended the play

Step-by-step explanation:

Step 1

Let the  number  of children be represented as  x

and the number of adults in the group  be represented as  as y

such that the total number of the group that went to the play

= x+y=11--- equation 1

And the total cost paid for the play will be expressed as

15x+  22y= 228------ equation 2

Step 2-- Solving

x+y=11--- equation 1

15x+  22y= 228------ equation 2

B y elimination method, multiply equation 1 by 15 to give equation 3  and then subtract from equation 2

15x+  22y= 228------ equation 2

15x+15Y= 165----- equation 3

7y=63

y= 63/7= 9

y= adults = 9

therefore x = number of children will be

x+y=11

x+ 9=11

x= 11-9= 2

In the group that attended the play,  Number of children = 2 and number of adults = 9

8 0
2 years ago
How do you write 11 thousands in standard form?
Afina-wow [57]
11,000 u think hope I helped
6 0
3 years ago
Use the standard normal distribution or the​ t-distribution to construct a 99​% confidence interval for the population mean. Jus
vichka [17]

Answer:

(25.732,30.868)

Step-by-step explanation:

Given that in a random sample of 42 ​people, the mean body mass index​ (BMI) was 28.3 and the standard deviation was 6.09.

Since only sample std deviation is known we can use only t distribution

Std error = \frac{s}{\sqrt{n} } =\frac{6.09}{\sqrt{42} } \\=0.9397

df = 42-1 =41

t critical for 99% two tailed = 2.733

Margin of error= 2.733*0.9397=2.568

Confidence interval lower bound = 28.3-2.568=25.732

Upper bound = 28.3+2.568=30.868

3 0
3 years ago
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