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bulgar [2K]
3 years ago
5

DUE TODAY PLEASE HELP​

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
7 0

Answer: 6. 113.398  7.  354.882  8. 4.82803

HOPE THIS HELPS!

Step-by-step explanation:

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7 divided by 2114 it is really hard
Ronch [10]

Answer:

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Step-by-step explanation:

3 0
3 years ago
The mathematics department of a college has 6 male professors, 9 female professors, 5 male teaching assistants, and 6 female tea
Hitman42 [59]

Given:

6 male professores

9 female professores

5 male teaching assistants

6 female teaching assistants

Sol:.

N(A\text{ or B)=N(A)+N(B)-N(A and B)}

N (professors)

\begin{gathered} =6+9 \\ =15 \end{gathered}

N(Male)

\begin{gathered} =6+5 \\ =11 \end{gathered}

N(professors and male)

=6

N(professors OR male) = N(professors) + N(males) -N(professor OR male)

\begin{gathered} N(\text{ Professors OR male)=15+11-6} \\ =20 \end{gathered}

N(People to choose from)

\begin{gathered} =6+9+5+6 \\ =26 \end{gathered}

Then probablitiy is:

\begin{gathered} =\frac{20}{26} \\ =\frac{10}{13} \end{gathered}

Then the probability is 10/13

6 0
1 year ago
Written as fractions, the decimal numbers 0.3 and 0.11 are 3/10 and 11/100, respectively
Vlada [557]
What exactly is the question here?
4 0
3 years ago
Read 2 more answers
how far would a 4.0 cm tall object be located from a reflecting surface with a focal lenght that is equal to 5 cm if it is to cr
katrin [286]

Answer:

HERE IS YOUR ANSWER

Step-by-step explanation:

Use the mirror equation:

1/di + 1/do = 1/f

where di = -10 cm and f = +15 cm. (Note that di is negative if the image is virtual.)

Substitute and solve for do.

1/do + 1/(-10 cm) = 1/(15 cm)

1/do = 1/(15 cm) - 1/(-10 cm) = 5/(30 cm)

do = 6 cm

Hope it helps you

Regards,

Rachana

5 0
3 years ago
Please Help!<br> Please write a sentence or more about the difference between​ each plot
topjm [15]
Box Plot has less variability in the data.  We can determine this by the distances between the beginning of the data to the end (range) and the distances between lower quartile and the upper quartile (interquartile range).

Box #1
Range - 30
IQR - 15

Box #2
Range - approximately 23
IQR - approximately 9

Box #2 has less variation in the data because the distances between these 2 ranges are smaller meaning the data is closer together.
6 0
3 years ago
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