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Mademuasel [1]
3 years ago
10

PLS ANYBODY HELP ME WITH MY GEOMETRY QUESTION

Mathematics
1 answer:
Ugo [173]3 years ago
6 0

Answer:

Angle A = 39°

Angle B = 90°

Angle C = 51°

Step-by-step explanation:

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A=24 and c=26<br>what does b=____<br>show work please​
Oksi-84 [34.3K]

Answer:

f jb gfbjfmnb fjb

Step-by-step explanation:

f bmf bknmf

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2 years ago
Round the following numbers to the nearest 10<br> a) 83<br> b) 127<br> c) 245
dolphi86 [110]
I believe the answer is-
a) 80
b) 130
c) 250
5 0
3 years ago
A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confid
katrin [286]

Answer:

The 95% confidence interval for the true proportion of all voters in the state who favor approval is (0.4384, 0.5050).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 865, \pi = \frac{408}{865} = 0.4717

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4717 - 1.96\sqrt{\frac{0.4717*0.5283}{865}} = 0.4384

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4717 + 1.96\sqrt{\frac{0.4717*0.5283}{865}} = 0.5050

The 95% confidence interval for the true proportion of all voters in the state who favor approval is (0.4384, 0.5050).

6 0
3 years ago
Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b
jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that  R should represent the set of all possible outcomes generated from  a bit string of length 10 .

So; as each place is fitted with either 0 or 1

\mathbf{|R|= 2^{10}}

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if  a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

\mathbf{P(E) = \dfrac{|E|}{|R|}}

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:

\mathbf{{|E|}=1 }   ; \mathbf{|R|= 2^{10}}

\mathbf{P(E) = \dfrac{1}{2^{10}}}

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3 years ago
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