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kondaur [170]
3 years ago
8

Pls help

Mathematics
1 answer:
denis-greek [22]3 years ago
7 0

Answer:

C I think

Step-by-step explanation:

Hehehehehehe

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PLEASEEEE HELP ME ILL GIVE YOU BRAINIEST!!
Jlenok [28]

Answer:

48 in³

<h3>        volume of triangular prism: Base Area * Height</h3>

Base Area:

\hookrightarrow \sf \dfrac{1}{2} *base*height

\hookrightarrow \sf \dfrac{1}{2} *3*8

\hookrightarrow \sf 12 \ in^2

Volume:

\hookrightarrow \sf 12 * 4

\hookrightarrow \sf 48 \ in^3

6 0
1 year ago
What is the value of (4-2)^3-3X4<br> o -20<br> o -4<br> o 4<br> o 20
hjlf

Answer:  -4 will be the answer

6 0
3 years ago
Read 2 more answers
Which values will complete the table.<br> A<br> B<br> C
Anvisha [2.4K]

Answer:

A=8

b=15

c=40

Step-by-step explanation:

5/8 as well as 10/16 = .625

a=8 blocks for 5 min given in question

b:24*0.625=15

c:25*40=.625 or 8 * 5=40 , since 5×5=25 taking 1st column times 5

3 0
3 years ago
In a clinical trial of 268 subjects treated with a​ drug, 11​% of the subjects reported dizziness. The margin of error is plus o
Mariulka [41]

Answer:

\hat p =0.11 represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

\hat q = 1-\hat p = 1-0.11= 0.89 represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

\alpha= 1-0.90 =0.1 and this value represent the significance level of the test or the probability of error type I

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

The margin of error is given by:

ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Lower= 0.11-0.04 = 0.07

Upper= 0.11+0.04 = 0.15

And for this case we have the following info:

\hat p =0.11 represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

\hat q = 1-\hat p = 1-0.11= 0.89 represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

\alpha= 1-0.90 =0.1 and this value represent the significance level of the test or the probability of error type I

4 0
3 years ago
A flat-screen television was marked down by 35 percent, which reduced its price by $115.65. What was the original cost of the te
Vera_Pavlovna [14]
Marked down by 35% this decreased it by 115.65 thhat means that 115.65 is 35% of original find original 35%=35/100=7/20 'of' means multiply 115.65=7/20 times original multiply both sides by 20/7 to clear fraction (20/7 times 7/20=140/140=1) 2313/7=original 330.428=original round 330.43 original price=$330.43
3 0
3 years ago
Read 2 more answers
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