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Alex777 [14]
3 years ago
9

8. The square of a number decreased by 3 times the number is 28. Find all

Mathematics
1 answer:
Llana [10]3 years ago
3 0
The answer is 7 i think
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I’m not sure if I got the answer right and I need help on the table I’m lost
harkovskaia [24]

The given information permits us to calculate the circumference (C) of the wheel as follows

C=\frac{265}{5}=53in

Then (as you said correctly!) the same wheel will move

(53in)\cdot16=848in

after 16 rolls.

3 0
1 year ago
Find the sequence of -1, 1, __, 5, __
marin [14]
The missing numbers are 3, and 7
the pattern is +2 to each number
7 0
2 years ago
ID: A
snow_tiger [21]

Answer:

13.63 miles per hour

Step-by-step explanation:

100 yards / 15 sec  converted to miles per hour

<u>100 yards</u> x <u> 60 sec</u>   x  <u> 60 min</u>   x  <u> 1 mile     </u>    =  13.63 miles/hr

   15 sec        1 min          1 hr           1760 yards

the mistake on Mrs Dukes is the conversion from miles to yards.

1 mile = 1760 yards NOT 5280 yards

7 0
3 years ago
23 – 7x2 + 19x – 19 is divided by 3 – 3?<br> result in the form q(x) + baby
snow_tiger [21]

Answer:

=−7x2+19x+ 41/3

Step-by-step explanation:

5 0
3 years ago
A study indicates that 37% of students have laptops. You randomly sample 30 students. Find the mean and the standard deviation o
Brilliant_brown [7]

Answer:

The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.

Step-by-step explanation:

Let <em>X</em> = number of students who have laptops.

The probability of a student having a laptop is, P (X) = <em>p</em> = 0.37.

A random sample of <em>n</em> = 30 students is selected.

The event of a student having a laptop is independent of the other students.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The mean and standard deviation of a binomial random variable <em>X</em> are:

\mu=np\\\sigma=\sqrt{np(1-p)}

Compute the mean of the random variable <em>X</em> as follows:

\mu=np=30\times0.37=1.11

The mean of the random variable <em>X</em> is 1.11.

Compute the standard deviation of the random variable <em>X</em> as follows:

\sigma=\sqrt{np(1-p)}=\sqrt{30\times0.37\times(1-0.37)}=\sqrt{0.6993}=0.836

The standard deviation of the random variable <em>X</em> is 0.836.

5 0
3 years ago
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