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Naya [18.7K]
3 years ago
9

Four students received the following grades on their math quizzes. Which student could use the mode to show that she is doing be

tter than the other three students?
Ashlee:78,94,90,70,94
Belen:79,79,100,95,99
Molly:66,73,98,88,98
Sofia:88,94,96,99,80

Answers choices:
a.Ashlee
b.Molly
c.Belen
d.Sofia
Mathematics
2 answers:
Ludmilka [50]3 years ago
7 0
Molly can use her mode to show that she is doing better than the other three students.

Ashlee’s mode is 94, Belen’s mode is 79, Molly’s is 98, and Sofia doesn’t have a mode. The highest is Molly, so she should use her mode to show she is doing better.
Elanso [62]3 years ago
6 0

Answer:

Ashlee’s mode is 94, Belen’s mode is 79, Molly’s is 98, and Sofia doesn’t have a mode

Step-by-step explanation:

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Answer:

I'm great actually I almost thought I had a bad grade in Spanish

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This table shows the number of points a high school basketball team scored during each of its last 10 games.
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The measure of central tendency that is most affected by the lowest score is; A: The mean.

<h3>How to find the measures of Central Tendency?</h3>

The points table is missing and so I have attached it.

The 3 main measures of central tendency are called mean, median and mode.

Now, from the table attached, if we find the average which is the mean, we will get 56.6.

Now, the mode is the number that occurs the most times. In this case, there is no mode because no number appears more than the other.

For the median, if we arrange in ascending order, we have;

42, 48, 49, 53, 55, 56, 58, 66, 68, 71.

The median = (55 + 56)/2 = 55.5

Now, if we remove the lowest score, the measure that will be most affected will be the mean as the median will only change by 0.5 to 56.

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3 0
2 years ago
Chanelle deposits $7,500 into the bank. She does not withdraw or deposit money for 6 years. She earns 6% interest during that ti
koban [17]

Answer: a. $2700

b. $10200

Step-by-step explanation:

a. The interest she would have earned at the end of the 6 years can be gotten using the formula

= PRT/100

= $7500 × 6% × 6

= $7500 × 6/100 × 6

= $7500 × 0.06 × 6

= $2700

b. Her balance when she wants to withdraw the money would be:

= $7500 + $2700

= $10200

5 0
3 years ago
Which term best describes the definition below?
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When a set of objects are chosen from a larger set in which the order of the object doesn't matter, we call this combination. Otherwise if the order of selection would matter this would be called a permutation.
Probability 
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4 0
4 years ago
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t

where, \text{R}_i_n = concentration of salt in the inflow \times input rate of brine solution

and \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

So, \text{R}_i_n = 4 lb/gal \times 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

                                                = 3 gal/min - 2 gal/min

                                                = 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

             = \frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min} = \frac{2A(t)}{500+t} \text{ lb/min }

Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;

= \frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }

or \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

4 0
3 years ago
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