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Vinvika [58]
3 years ago
15

The middle school is 2.72 km from Marsha's house and 1.54 km from Ryan's house. How much farther does Marsha live from the middl

e school then Ryan
Mathematics
1 answer:
AlladinOne [14]3 years ago
3 0
Ryan lives 1.54 km from middle school
Marsha lives 2.72 km from middle school

2.72 - 1.54 = 1.18....Marsha lives 1.18 km further 
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large cheese pizzas cost $5 each and large one topping pizzas cost $6 each. write an equation that represents the total cost, T
Gelneren [198K]

Answer:

T = 5c +5d,    where unit of the equation is $.

Step-by-step explanation:

Cost of 1 large cheese  pizza = $5

No. of large cheese pizza = c

therefore

cost of c  large cheese pizza = c*Cost of 1 large cheese  pizza = $5c

Cost of 1 large one topping  pizza = $6

No. of large one topping pizza = d

therefore

cost of d  large one topping pizza = d*Cost of 1 large one topping  pizza =             cost of d  large one topping pizza = $6d  

Total cost of  c  large cheese pizza and cost of d  large one topping pizza

= $5c +$6d

total cost is represented by T

thus, T = $5c +$5d  Answer

4 0
3 years ago
Ken wants to sell some of the baby angelfish when they are old enough. He wants to sell 24 of the angel fish for 40 cents each.
Nina [5.8K]

Answer:

Step-by-step explanation:

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6 0
3 years ago
( x + 9 ) - ( x -2 )=
Triss [41]

Answer:

x = -7/2

Step-by-step explanation:

( x + 9 ) - ( x -2 )=

2x +7 =

2x = -7

x = -7/2

5 0
2 years ago
IQ scores on the WAIS test approximate a normal curve with a mean of 100 and a standard deviation of 15. What IQ score is identi
riadik2000 [5.3K]

Answer:

IQ scores of at least 130.81 are identified with the upper 2%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 100 and a standard deviation of 15.

This means that \mu = 100, \sigma = 15

What IQ score is identified with the upper 2%?

IQ scores of at least the 100 - 2 = 98th percentile, which is X when Z has a p-value of 0.98, so X when Z = 2.054.

Z = \frac{X - \mu}{\sigma}

2.054 = \frac{X - 100}{15}

X - 100 = 15*2.054

X = 130.81

IQ scores of at least 130.81 are identified with the upper 2%.

7 0
3 years ago
ACT scores have a mean of 20.8 and 9 percent of the scores are above 28. The scores have a distribution that is approximately no
dimaraw [331]

Answer: 5.37

Step-by-step explanation:

Let x =  ACT scores.

Given: ACT scores have a mean of 20.8 and 9 percent of the scores are above 28. The scores have a distribution that is approximately normal.

i.e. P(X>28)=0.09   (i)

Now,

P(X>28)=P(\dfrac{X-\mu}{\sigma}>\dfrac{28-20.8}{\sigma})\\\\= P(z>\dfrac{7.2}{\sigma})\ \ \ \ [z=\dfrac{X-\mu}{\sigma}]    (ii)

One -tailed z value for p-value of 0.09 =1.3408 [By z-table]

From (i) and (ii)

\dfrac{7.2}{\sigma}=1.3408\\\\\Rightarrow\ \sigma=\dfrac{7.2}{1.3408}\\\\\Rightarrow\ \sigma=5.37

Hence, the standard deviation = 5.37

5 0
3 years ago
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