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yaroslaw [1]
3 years ago
6

Please help thank you so much

Mathematics
2 answers:
kirza4 [7]3 years ago
8 0

Answer: 15

Step-by-step explanation:

Since the angle looks to be around 90 degrees and the x is timsed by 6 the answer would have to be a lower number. And from the options only 15 would work.

Licemer1 [7]3 years ago
5 0

Answer:

\Huge\boxed{x=15}

Step-by-step explanation:

Hello There!

The angle labeled 6x is a right angle

If an angle is formed using the diameter of a circle then the opposite angle of that side is a right angle

So 6x = 90

isolate the variable

divide each side by 6

90/6= 15

6x/6=x

we're left with x = 15

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6х + 4y + 2z = 32<br>3х - 3y - z = 19<br>3х + 2y + z = 32<br>x = ?<br>y = ?<br>z = ?​
blsea [12.9K]

×=12

y=9

z=2 not sure I'm right

3 0
3 years ago
1. Find f(3) for<br> the function<br> below.<br> f(x) = 4^x
Wewaii [24]

Answer:

64

Step-by-step explanation:

f(x) = 4^x,

When x = 3,

=> f(3) = 4^3

= 4 * 4 * 4

= 64

7 0
3 years ago
Read 2 more answers
Advertisements for the Sylph Physical Fitness Center claim that completion of their course will result in a loss of weight (meas
MrMuchimi

Answer:

90% confidence interval for the mean weight loss for the population represented by this sample assuming that the differences are coming from a normally distributed population is [1.989 ,13.5105]

Step-by-step explanation:

The data given is

                        Mean                 Std Dev

Before            177.25                29.325

After              169.50                    22.431

Difference        7.75                    8.598

Hence d`=   7.75 and sd=   8.598

The 90% confidence interval for the difference in means for the paired observation is given by

d` ± t∝/2(n-1) *sd/√n

Here  t∝/2(n-1)=1.895  where n-1= 8-1= 7 d.f

and  ∝/2= 0.1/2=0.05

Putting the values

d` ± t∝/2(n-1) *sd/√n

7.75  ±1.895 *  8.598 /√8

 7.75  ± 5.7605

1.989 ,13.5105

90% confidence interval for the mean weight loss for the population represented by this sample assuming that the differences are coming from a normally distributed population is [1.989 ,13.5105]

3 0
3 years ago
Need help with this math don’t understand
slava [35]

Answer:

A

Step-by-step explanation:

The formula for flipping smth over the x axis is(x,y)→(x,−y)

just flip the symbol on the second #

your points are:

(-4,-2)→(-4,2)

(0,-4)→(0,4)

(-4,-5)→(-4,5)

Once you graph it, youll see that the shapes are congruent, just flipped, therefore the answer is A

3 0
3 years ago
Average box of crackers is 24.5 ounces with standard deviation of. 8 ounce. What percent of the boxes weigh more than 22.9 ounce
34kurt

Answer:

97.7% of of the boxes weigh more than 22.9 ounces.

15.9% of of the boxes weigh less than 23.7 ounces.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  24.5 ounces

Standard Deviation, σ = 0.8 ounce

We are given that the distribution of boxes weight is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(boxes weigh more than 22.9 ounces)

P(x > 22.9)

P( x > 22.9) = P( z > \displaystyle\frac{22.9 - 24.5}{0.8}) = P(z > -2)

= 1 - P(z \leq -2)

Calculation the value from standard normal z table, we have,  

P(x > 22.9) = 1 - 0.023 =0.977= 97.7\%

97.7% of of the boxes weigh more than 22.9 ounces.

b) P(boxes weigh less than 23.7 ounces)

P(x < 23.7)

P( x < 23.7) = P( z < \displaystyle\frac{23.7 - 24.5}{0.8}) = P(z < -1)

Calculation the value from standard normal z table, we have,  

P(x < 23.7) =0.159= 15.9\%

15.9% of of the boxes weigh less than 23.7 ounces.

7 0
4 years ago
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