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anzhelika [568]
3 years ago
14

I need help solving this question please :)

Mathematics
1 answer:
dsp733 years ago
3 0

Answer:

um can you pleasee take a clearer picture thankyou I cant see the things inside the circle properly

Step-by-step explanation:

You might be interested in
The cost of performance tickets and beverages for a family of four can be modeled using the equation 4x + 12 = 48, where x
zhenek [66]

Answer:

The cost of one ticket is $9

Step-by-step explanation:

we have

4x+12=48

where

x represents the cost of a ticket

Solve for x

Subtract 12 both sides

4x=48-12

4x=36

Divide by 4 both sides

x=36/4=9

therefore

The cost of one ticket is $9

4 0
3 years ago
-1/2+c=31/4<br><br> A. C=33/4<br> B. C=8<br> C. 29/4<br> D. 7
kkurt [141]

Answer:

A

Step-by-step explanation:

-1/2+c=31/4

-1/2+c+1/2=31/4+1/2

c=33/4

4 0
3 years ago
Anyone know how to prove these are equal to one another? I also need all the steps.
IrinaVladis [17]

Answer:

Step-by-step explanation:

it can be simplified to \frac{sec \theta}{cot\theta+tan\theta} =sin\theta  . if you multiply each side by the denominator, it becomes sec\theta=\frac{sin^2\theta}{cos\theta} + cos\theta since tan\theta=\frac{sin\theta}{cos\theta} and cotan is the reciprocal of that. Because sec\theta=\frac{1}{cos\theta} so multiplying each side gives 1=sin^2\theta+cos^2\theta, and sin^2\theta+cos^2\theta is equal to 1.

6 0
1 year ago
10.
Kay [80]

Answer:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.535 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.2 \leq \sigma \leq 2.8

And the best option would be:

A.  2.2 < σ < 2.8

Step-by-step explanation:

Information provided

\bar X=32.1 represent the sample mean

\mu population mean  

s=2.4 represent the sample standard deviation

n=83 represent the sample size  

Confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom are given by:

df=n-1=83-1=82

The Confidence is given by 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, the critical values for this case are:

\chi^2_{\alpha/2}=62.132

\chi^2_{1- \alpha/2}=104.139

And replacing into the formula for the interval we got:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.535 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.2 \leq \sigma \leq 2.8

And the best option would be:

A.  2.2 < σ < 2.8

3 0
3 years ago
Lisa and 3 of her friends want to share 3 apples equally. write an expression to show what fraction of the apples each person sh
svp [43]

1/3 is your answer

2/6 and 3/9 are equivalent

4 0
3 years ago
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