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3 years ago

Does 23^-1 (mod 1000) exist? If yes solve it.

Mathematics

1 answer:

sweet [91]3 years ago

3 0

Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).

Let x be the inverse. Then x is such that

23x ≡ 1 (mod 1000)

Use the Euclidean algorithm to solve for x :

1000 = 43×23 + 11

23 = 2×11 + 1

→ 1 ≡ 23 - 2×11 (mod 1000)

→ 1 ≡ 23 - 2×(1000 - 43×23) (mod 1000)

→ 1 ≡ 23 - 2×1000 + 86×23 (mod 1000)

→ 1 ≡ 87×23 - 2×1000 ≡ 87×23 (mod 1000)

→ 23⁻¹ ≡ 87 (mod 1000)

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A marble is randomly selected from a bag containing 15 black, 12 white, and 6 clear marbles. Find P(not clear). Round to the nea

dexar [7]

15+12+6= total number of marbles
33= total
Not clear marbles = 15+12 or 27
27 /33
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82 percent

8 0

3 years ago

Use the given minimum and maximum data entries, and the number of classes, to find the class width, the lower class limits,

Gennadij [26K]

Using proportions and the information given, it is found that:

The class width is of 14.375.

The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.

The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

-------------------------

Minimum value is 19.
Maximum value is of 134.
There are 8 classes.
The classes are all of equal width, thus the width is of:

$W = \frac{134 - 19}{8} = 14.375$

-------------------------

The intervals will be of:

19 - 33.375

33.375 - 47.750

47.750 - 62.125

62.125 - 76.500

76.500 - 90.875

90.875 - 105.250

105.250 - 119.625

119.625 - 134.

The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.

The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

A similar problem is given at brainly.com/question/16631975

6 0

3 years ago

A=1/2bh. A=200ft. H=10. What is b

Svetllana [295]

Answer:

B= $\frac{2A}{H}$

Step-by-step explanation:

If you noticed, A= $\frac{1}{2}$ BH is the formula for finding the area for a triangle. Your goal is to get B by it's self. Your first step will be to clear of the fraction first, so you will multiply both sides by 2. 2(A)=2( $\frac{1}{2}$ BH). On the left, you have 2×A= On the right side, you have 2( $\frac{1}{2}$ BH), but since you have a number in the equation, you will only use 2× $\frac{1}{2}$ . To solve 2× $\frac{1}{2}$ , you will cnacel out both 2's and you have 1. 1×BH will still equal BH, so you are now left will B×H.

(Your new equation looks like this by the way). 2A=BH

Since you need to get B by its self, the way to clear the H away from the B is by dividing. You will now divide the B and H aswell as 2A and H. (It will look like this) $\frac{2A}{H} = \frac{BxH}{H}$ . (Again when you have the same number or letter, you cross it out. When you divide, you won't change anything on the left side, and all you have to do on the right id to cross out the H next to the B and cross out the H on the bottom of the equation). You should be left with $\frac{2A}{H}$ = B. Now you can turn it around for your final answer. B= $\frac{2A}{H}$ .

Please let me know if i helped, how I did, and if you have any questions.

3 0

3 years ago

The weights of bags of ready-to-eat salad are normally distributed with a mean of 290 grams and a standard deviation of 10 grams

pentagon [3]

290-10=280

280 is one standard deviation below.
All that are above mean is 50%.
One standard deviation below is 34%.
50%+34%=84% all of these bags weigh more than 280 g.
So,
100%-84%=16% should be bags that are less than 280 g.

Answer is b) 16 %.

5 0

3 years ago

A candle maker has rectangular block of paraffin that is 4 centimeter high by 8 centimeter long by 6 centimeter wide. He melts i

Veronika [31]

Answer:

The candle will not use all the paraffin from the original block.

Step-by-step explanation:

The cone-shaped candle is 5 cm wide at the base i.e. the diameter of the base is 5 cm i.e. radius is 2.5 cm.

If the height of the cone is 25 cm, then total volume of the cone shaped candle is

$\frac{1}{3}\pi r^{2} h$

= $\frac{22 \times 2.5^{2} \times 25 }{3 \times 7}$

= 163.69 cubic cm.

Again, the volumes of the rectangular block of paraffin is (4 × 8 × 6) = 192 cubic cm.

Since, 192 > 163.69

Therefore, the candle will not use all the paraffin from the original block.

5 0

3 years ago