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kow [346]
3 years ago
14

P divided by 5 equals -32

Mathematics
2 answers:
Flauer [41]3 years ago
7 0
The answer is p=-160
Lostsunrise [7]3 years ago
6 0

Answer:

p = -160

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

Step-by-step explanation:

<u>Step 1: Define</u>

p/5 = -32

<u>Step 2: Solve for </u><em><u>p</u></em>

  1. Multiply 5 on both sides:                     p = -160
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From online resources gotten about this question, for quadrilateral EFGH and quadrilateral E'F'G'H to be congruent, what we must do first is to rotate 180° counterclockwise around the origin and then move EFGH onto E'F'G'H'.

The last step to get this proof of congruency is to reflect across the y-axis.

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Assume that in a particular military exercise involving two units, Red and Blue, thereis a 60 % chance that the Red unit will su
Marysya12 [62]

Answer: a) There are 42% chances that both units will meet their objectives.

b) There are 88% chances that one or the other but not both of the units will be successful.

Step-by-step explanation:

Since we have given that

Probability that the Red unit will successfully meet its objectives = 60% = P(R)

Probability that Blue unit will successfully meet its objectives = 70% = P(B)

Probability that only Red unit will be successful = P(only Red) = 18%

As we know that

P(only\ red)=P(R)-P(R\cap B)\\\\0.18=0.60-P(R\cap B)\\\\0.18-0.60=-P(R\cap B)\\\\-0.42=-P(R\cap B)\\\\P(R\cap B)=42\%

Hence, there are 42% chances that both units will meet their objectives.

the probability that one or the other but not both of the units will be successful is given by

P(R\cup B)=P(R)+P(B)-P(R\cap B)\\\\P(R\cup B)=0.60+0.70-0.42\\\\P(R\cup B)=0.88\\\\P(R\cup B)=88\%

Hence, there are 88% chances that one or the other but not both of the units will be successful.

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