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zhenek [66]
3 years ago
14

PLEASE HELP ASAP The skate pro company manufactures skateboards. They found 12 defective skateboards in a batch of 400. How many

defective skateboards might they find in a batch of 1,200?
What is the probability that a skateboard is defective?SIMPLIFY
a)3/100
b)12/400
c)4/100
Mathematics
2 answers:
pishuonlain [190]3 years ago
5 0
A is the answer I think have good day
horsena [70]3 years ago
4 0

Answer:

I believe the answer is A. love your pfp btw

Step-by-step explanation:

12/400 is .03 or 3%

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is this for the test on schoolobjects?

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<em>Hi there !</em>

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What is the solution to the equation 2 (x-3) ^2 =13 help help ASAP !
olganol [36]

Answer:

3+sqrt(13/2)

3-sqrt(13/2)

Step-by-step explanation:

2 (x-3) ^2 =13

Divide each side by 2

2/2 (x-3) ^2 =13/2

(x-3) ^2 =13/2

Take the square root of each side

sqrt((x-3) ^2) =±sqrt(13/2)

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x-3+3 = 3±sqrt(13/2)

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4 0
3 years ago
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A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student
yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means is:

CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}

The information provided is as follows:

n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43

The critical value of <em>z</em> for 98% confidence level is,

z_{\alpha/2}=z_{0.02/2}=2.326

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}

     =(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0
3 years ago
What's the value of x in the equation: 10x-6=5x34
Snowcat [4.5K]

Answer:

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// have a great day //

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