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ASHA 777 [7]
2 years ago
10

Help help help help help helppp!!!

Mathematics
1 answer:
kkurt [141]2 years ago
6 0

Answer:

kids = 12; adults = 8

Step-by-step explanation:

12 x $4 = $48

$6 x 8 = $48

$48 + $48 = $96

I hope this helps!

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Clint needs to decrease his budget for groceries by 5% each month. He currently allots for $500 a month on groceries. What will
Radda [10]

Answer:

$386.9

Step-by-step explanation:

Given that current grocery costs are $500 and he want to reduce 5% every month

We can create a geometric progression with a = 500 and common ratio(r) = 0.95

because every time we reduce 5% we actually make the actual cost as 95%

T(n)=ar^{n-1}

Now we should find n=6

T(6)=500\times(0.95)^{6-1}

t(6)=386.9

The allotment is $386.9

8 0
3 years ago
The amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship co
zvonat [6]

Answer:

95% Confidence interval for σ2 and for σ is (3.33 , 38.85) and (1.82 , 6.23) respectively.

Step-by-step explanation:

We are given that the amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.83 mils.

Assuming data follows normal distribution.

So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;

        P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

where, s = sample standard deviation = 2.83 mils

          \sigma^{2} = population variance

           \sigma = population standard deviation

           n = sample size = 7

<em>So, 95% confidence interval for population variance, </em>\sigma^{2} <em>is;</em>

P(1.237 < \chi^{2} __n_-_1 < 14.45) = 0.95 {As the table of at 6 degree of freedom

                                                     gives critical values of 1.237 & 14.45}

P(1.237 < \frac{(n-1)s^{2} }{\sigma^{2} } < 14.45) = 0.95

P( \frac{ 1.237}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{ 14.45}{(n-1)s^{2} } ) = 0.95

P( \frac{ (n-1)s^{2}}{14.45 } < \sigma^{2} < \frac{ (n-1)s^{2}}{1.237 } ) = 0.95

95% confidence interval for \sigma^{2} = ( \frac{ (n-1)s^{2}}{14.45 }  , \frac{ (n-1)s^{2}}{1.237 }  )

                                                  = ( \frac{ (7-1) \times 2.83^{2}}{14.45 } , \frac{ (7-1) \times 2.83^{2}}{1.237 } )

                                                  = (3.33 , 38.85)

95% C.I. for population standard deviation, \sigma  = ( \sqrt{3.33} , \sqrt{38.85} )

                                                                            = (1.82 , 6.23)

Therefore, 95% confidence interval for the population variance (σ2) and population standard deviation (σ) are (3.33 , 38.85) and (1.82 , 6.23) respectively.

7 0
3 years ago
20 POINTS What is the area of this triangle?
Nikitich [7]

Answer:

46.1

Step-by-step explanation:

Mark BRAINLIEST

6 0
3 years ago
Read 2 more answers
On a 6-sided die, what is the probability of rolling an even number that is not 4? 1/6 1/3 2/3 5/6
myrzilka [38]
The probability of rolling an even number on a six sided die is 1/3. 
8 0
4 years ago
Write the equation of the line in slope-intercept form that has the following points: (1, -3)(0, -5)
Stolb23 [73]

Answer:

A. y = 2x - 5

Step-by-step explanation:

8 0
4 years ago
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