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lianna [129]
3 years ago
13

On marys walk, she covered the 3 mile distance in one hour. However, the return trip took an hour and a half. What was her avera

ge speed?
Mathematics
2 answers:
Lana71 [14]3 years ago
6 0

Answer:

2.4 mph

Step-by-step explanation:

lesya [120]3 years ago
4 0

Answer:

Step-by-step explanation:

Speed = distance/time

Speed= 3/1= 3miles/hour

Return trip

Speed=3miles/1.5hour=2miles/hour

Average speed= initial + final trip= 2miles/hour+3miles/hour, then divide by 2

= 5/2=2.5miles/hour

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If two lines are parrallel do they have the same slope
victus00 [196]

Answer:

no i dont think so becuse they can still be angeled in all diffrent directions

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
tickets for a football match are sold at $30 for adults and $15 for children a company bought 28 tickets if x of these tickets w
anzhelika [568]
First, we must let:
   x = number of tickets intended for adults
   y = number of tickets intended for children.

a. Write in terms of x the number of tickets for children
     Solution:
           x + y = 28   ⇔    y = 28 - x   (equation 1)
       To answer in terms of x:
           no. of tickets for tickets for children = 28 - x

b. the amount spent on tickets for adults
    Solution:  $30 is the cost of ticket per adult and there are x number of tickets intended for adults.
           Therefore, 
            amount spent on ticket for adults = 30x

c. the amount spent on the tickets.
     Solution:
       $ 15  = cost of ticket per child
       $ 30 = cost of ticket per adult

      total amount spent on tickets = 30x + 15y    ⇒   (equation2)
  substitute equation 1 to equation 2.
  (equation 1)   y = 28 - x
  (equation 2)   total amount spent on tickets = 30x + 15y
                        total amount spent on tickets = 30x + 15(28-x)
                        total amount spent on tickets = 30x + 420 - 15x
                        total amount spent on tickets = 15x + 420
4 0
3 years ago
Which set of angles can form a triangle
kow [346]
Answer is c explanation
7 0
4 years ago
Read 2 more answers
The greatest factor that two or more numbers have in common is the
trapecia [35]

Answer:

 "greatest common factor" (GCF) or "greatest common divisor" (GCD)

Step-by-step explanation:

Apparently, you're looking for the term that has the given definition. It is called the GCF or GCD, the "greatest common factor" or the "greatest common divisor."

_____

The GCF or GCD can be found a couple of ways. One way is to find the prime factors of the numbers involved, then identify the lowest power of each of the unique prime factors that are common to all numbers. The product of those numbers is the GCF.

<u>Example</u>:

  GCF(6, 9)

can be found from the prime factors:

  • 6 = 2·3
  • 9 = 3²

The unique factors are 2 and 3. Only the factor 3 is common to both numbers, and its lowest power is 1. Thus ...

  GCF(6, 9) = 3¹ = 3

__

Another way to find the GCD is to use Euclid's Algorithm. At each step of the algorithm, the largest number modulo the smallest number is found. If that is not zero, the largest number is replaced by the result, and the process repeated. If the result is zero, the smallest number is the GCD.

  GCD(6, 9) = 9 mod 6 = 3 . . . . . (6 mod 3 = 0, so 3 is the GCD)

8 0
3 years ago
In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi
Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n  = 50

sample mean \overline x = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

X \sim N ( \mu \sigma)

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})

P(X>140) = 1 - P( Z\leq0.42)

From z tables,

P(X>140) = 1 - 0.6628

P(X>140) = 0.3372

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

Y \sim Binomial (np)

Y \sim Binomial (50,0.3372)

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49}  + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +...  + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}

P(Y ≥ 15) = 0.763

7 0
3 years ago
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