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tigry1 [53]
2 years ago
14

This chart shows the number of books that eight students read last year.

Mathematics
2 answers:
erastova [34]2 years ago
6 0

Answer:

B is the correct answer I believe

malfutka [58]2 years ago
6 0

Answer:

A.

Step-by-step explanation:

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name the set of 6 consecutive integers starting with -3 put the set in braces and put commas between the elements of the set
melisa1 [442]
6-3=3 that is the answer ok
6 0
3 years ago
Help me please i'm confused again
11Alexandr11 [23.1K]

Answer: 6¹⁰

Step-by-step explanation:

\displaystyle\\6^4*6^6=\\\\6^{4+6}=\\\\6^{10}

6 0
1 year ago
I need help with this please and thank you!
Vlad [161]

Answer:  2x^2      -16t

               25t         -20

Step-by-step explanation:

Multiply the corresponding terms to find out which answer belongs in each box. For the top left box you would multiply  5t x 4t = 20t^2. For the top right multiply -4 x 4t = -16t. For bottom left box multiply 5t x 5 = 25t and for the bottom right box multiple -4 x 5 = -20.

6 0
3 years ago
A van is traveling 1.5 km per minute. How fast is the van traveling in miles per hour?
ipn [44]

Answer:

55.9234

Step-by-step explanation:

1.5*60=90

90km=(x) miles

3 0
2 years ago
Read 2 more answers
The measurement of the height of 600 students of a college is normally distributed with a mean of
ioda

Answer:

68

Step-by-step explanation:

We let the random variable X denote the height of students of the college. Therefore, X is normally distributed with a mean of 175 cm and a standard deviation of 5 centimeters.

We are required to determine the percent of students who are between 170 centimeters and 180 centimeters in height.

This can be expressed as;

P(170<X<180)

This can be evaluated in Stat-Crunch using the following steps;

In stat crunch, click Stat then Calculators and select Normal

In the pop-up window that appears click Between

Input the value of the mean as 175 and that of the standard deviation as 5

Then input the values 170 and 180

click compute

Stat-Crunch returns a probability of approximately 68%

5 0
3 years ago
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