PLEASE HELP ANYONE PLEASE HELP !!!!

What is the minimum number of binary bits needed to represent each of the following unsigned decimal integers? a. 65

professor190 [17]

Answer: N >/= 7 bits

Minimum of 7 bits

Step-by-step explanation:

The minimum binary bits needed to represent 65 can be derived by converting 65 to binary numbers and counting the number of binary digits.

See conversation in the attachment.

65 = 1000001₂

65 = 7 bits :( 0 to 2^7 -1)

The number of binary digits is 7

N >/= 7 bits

7 0

3 years ago

Read 2 more answers

H e l p. P l e a s e e e

horrorfan [7]

Let's choose 43° to be our angle. We know the opposite, and want to know the hypotenuse. Sine is the trig. we use for opposite over hypotenuse.
sinθ = opposite/hypotenuse
sin(43)=27/x
We want to isolate <em>x</em>, so that looks like:
x = 27/sin(43)
Therefore, the value of x is 27/0.682 or ≈39.589 or F

4 0

3 years ago

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Use the image above to write a conjecture about regular polygons and lines of symmetry

Tatiana [17]

To find conjecture about regular polygon, use the formula $\frac{360}{n}$

Where, n = number of sides of the regular polygone

1) Triangle
n =3
So, Conjecture of the triangle = $\frac{360}{3}$ = 120

2) Square
n =4
So, Conjecture of the square = $\frac{360}{4}$ = 90

3) Pentagone
n = 5
So, Conjecture of the pentagon = $\frac{360}{5}$ = 72

4) Hexagone
n = 6
So, Conjecture of the pentagon = $\frac{360}{6}$ = 60.

7 0

4 years ago

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In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

How do you work this out? We have to make one positive and the other negative then put it on a graph correctly

ANEK [815]

The answer to Q1 along with an explanation is shown in the picture;
Having a look at all the questions, they are all essentially the same type so can all be solved similarly;
So by following the method and explanation, you should be able to do the rest of the questions.

7 0

3 years ago