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3 years ago

In this figure, AB∥CD and m∠1=110°. What is m∠5 ? 90° 180° 70° 110°

Mathematics

1 answer:

WINSTONCH [101]3 years ago

6 0

Answer:

110 hope u get a good grade:) add me on ig welia.martinez

Step-by-step explanation:

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Work out m and c for the line: y−4x=−1

Eduardwww [97]

Step-by-step explanation:

y -4x = -1

y = 4x -1

then y = Mx + b

m = 4

b = -1

7 0

3 years ago

4. Steve has a card collection. He keeps 12 of the cards in a pocket, which

BabaBlast [244]

Answer:

I think 80

Step-by-step explanation:

100/15=6.66666667

12x6.67=80.4

4 0

3 years ago

A grocer wants to make a 10-pound mixture of peanuts and cashews that he can sell for $4.75 per pound. If peanuts cost $4.00 per

laiz [17]

Answer:

3lbs of Cashews

Step-by-step explanation:

lbs of Cashews, and 7 lbs of Peanuts

$4.00P + 6.50C = ($4.75/lbs)(10lbs)

$4.00(7) + $6.50(3) = $47.50

$28.00 + $19.50 = $47.50

$47.50 = $47.50

Therefore it's 3lbs of Cashews

8 0

3 years ago

Read 2 more answers

The manager of a music store has kept records of

Marysya12 [62]

Answer:

(a) $P(x\le 3) = 0.75$

(b) $P(x\le 3) = 0.75$

(b) is the same as (a)

(d) $P(x = 1\ or\ 2) = 0.55$

(e) $P(x > 2) = 0.45$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

Solving (b): Probability of at most 3 CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

(b) is the same as (a)

Solving (c): Probability of 5 or more CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x \ge 5) = P(5) + P(6\ or\ more)$

This gives:

$P(x\ge 5) = 0.05 + 0.05$

$P(x \ge 5) = 0.10$

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}$

This probability is calculated as:

$P(x = 1\ or\ 2) = P(1) + P(2)$

This gives:

$P(x = 1\ or\ 2) = 0.30 + 0.25$

$P(x = 1\ or\ 2) = 0.55$

Solving (e): Probability of more than 2 CDs

Here, we consider:

$\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)$

This gives:

$P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05$

$P(x > 2) = 0.45$

3 0

3 years ago

B + d = -2d + a solve for b

77julia77 [94]

I believe that it is b=-3d+a

5 0

3 years ago