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svlad2 [7]
4 years ago
14

Can someone please help me with this. please & thank you !

Mathematics
1 answer:
eduard4 years ago
7 0

Answer:

C. 62

Step-by-step explanation:

because all of the angles must = to 180 degrees and the 2 bottom must be equivalent so 62+62+56=180

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(4m+7n-6p)-(2m-3n-4p)
Vanyuwa [196]

Answer:

2m + 10n - 2p

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Distributive Property

<u>Algebra I</u>

  • Terms

Step-by-step explanation:

<u>Step 1: Define</u>

(4m + 7n - 6p) - (2m - 3n - 4p)

<u>Step 2: Simplify</u>

  1. Distribute negative:                    4m + 7n - 6p - 2m + 3n + 4p
  2. Combine like terms (m):             2m + 7n - 6p + 3n + 4p
  3. Combine like terms (n):              2m + 10n - 6p + 4p
  4. Combine like terms (p):              2m + 10n - 2p
6 0
3 years ago
Given quadrilateral MATH is similar to quadrilateral ROKS calculate the value of MH Picture is below
tino4ka555 [31]
<h3>Answer:  MH = 7</h3>

=====================================================

Explanation:

The double tickmarks for quadrilateral MATH show that MA = TH. Since TH is 5 units long, this makes MA the same length as well.

For quadrilateral ROKS, we have RO = 15. For "MATH" and "ROKS" we have "MA" and "RO" as the first two letters of each four-letter sequence; meaning that MA and RO correspond together.

The ratio of the corresponding segments is RO/MA = 15/5 = 3.

The larger quadrilateral has each side length 3 times longer than the smaller quadrilateral's corresponding side lengths.

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

In short,

larger side = 3*(smaller side)

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

Using this scale factor of 3, we can find MH

larger side = 3*(smaller side)

RS = 3*(MH)

21 = 3*MH

3*MH = 21

MH = 21/3

MH = 7

3 0
4 years ago
ILLL markk brainliest i promise plsss helpp
SOVA2 [1]

Answer:

3

Step-by-step explanation:

8 0
4 years ago
Candy. Someone hands you a box of a dozen chocolate-covered candies, telling you that half are vanilla creams and the other half
BaLLatris [955]

Answer:

a) P=0.091

b) If there are half of each taste, picking 3 vainilla in a row has a rather improbable chance (9%), but it is still possible that there are 6 of each taste.

c) The probability of picking 4 vainilla in a row, if there are half of each taste, is P=0.030.

This is a very improbable case, so if this happens we would have reasons to think that there are more than half vainilla candies in the box.

Step-by-step explanation:

We can model this problem with the variable x: number of picked vainilla in a row, following a hypergeometric distribution:

P(x=k)=\dfrac{\binom{K}{k}\cdot \binom{N-K}{n-k}}{\binom{N}{n}}

being:

N is the population size (12 candies),

K is the number of success states in the population (6 vainilla candies),

n is the number of draws (3 in point a, 4 in point c),

k is the number of observed successes (3 in point a, 4 in point c),

a) We can calculate this as:

P(x=3)=\dfrac{\binom{6}{3}\cdot \binom{12-6}{3-3}}{\binom{12}{3}}=\dfrac{\binom{6}{3}\cdot \binom{6}{0}}{\binom{12}{3}}=\dfrac{20\cdot 1}{220}=0.091

b) If there are half of each taste, picking 3 vainilla in a row has a rather improbable chance (9%), but is possible.

c) In the case k=4, we have:

P(x=3)=\dfrac{\binom{6}{4}\cdot \binom{6}{0}}{\binom{12}{4}}=\dfrac{15\cdot 1}{495}=0.030

This is a very improbable case, so we would have reasons to think that there are more than half vainilla candies in the box.

4 0
4 years ago
I need help please....
zhuklara [117]
The answer is C!!!  I really hope that this helps you
6 0
3 years ago
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