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aleksley [76]
2 years ago
6

The frequency table shows the self-reported number of minutes students spent studying for a statistics test. What percent of the

students reported studying between 50 and 59 minutes.

Mathematics
1 answer:
tankabanditka [31]2 years ago
4 0

Answer:

15.38 of the students reported studying between 50 and 59 minutes.

Step-by-step explanation:

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Simplify. 4−16÷4+42 −16 13 15 16
Dimas [21]

Answer:

[See Below]

Step-by-step explanation:

✦ First split the equation in little pieces and solve them:

  ✧ 4 - 16 = -12

  ✧ 4 + 4² = 20 (4² = 16)

✦ Now divide them:

  ✧ -12 ÷ 20 = -0.6

So I'm guessing you're trying to simplify 4²... so it'd be 16 because none of the choices are the simplified version of the problem.

~<em>Hope this helps Mate. If you need anything feel free to message me.</em>

7 0
3 years ago
Please could someone help with this ? Thankyou
katovenus [111]

Answer:

1 2/7

Step-by-step explanation:

4.5:3 1/2

45/10 ÷ 7/2

45/10 x 2/7

90/70=9/7

=1 2/7

4 0
3 years ago
Read 2 more answers
Which one has a greater value, , -7 + (-7) or -7 - (-7)? Explain your answer. GIVING BRAINLIEST!!!
dmitriy555 [2]
-7 -(-7) because it will equal zero and the other one will be -14
6 0
3 years ago
Write an expression to represent: the product of 5 and a number decreased by 3.
Maru [420]

Answer:

(n) (5) - 3

5n - 3

7 0
3 years ago
Solve <br>In 2x + In 2=0​
irakobra [83]

Solving  ln(2x)+ln(2)=0 we get, x=\frac{1}{4}

Step-by-step explanation:

We need to solve: ln(2x)+ln(2)=0

Solving:

ln(2x)+ln(2)=0

Adding -ln(2) on both sides:

ln(2x)+ln(2)-ln(2)=0-ln(2)

ln(2x)=-ln(2)

Using logarithmic rule: if log_a b =c\,\,then\,\, b=a^c

So,

2x=e^{-ln(2)}

Simplifying:

2x=(e^{ln(2)})^{-1}\\e^{ln(2)}=2\\2x=(2)^{-1}\\2x=\frac{1}{2} \\x=\frac{1}{2x2}\\x=\frac{1}{4}

So, Solving  ln(2x)+ln(2)=0 we get, x=\frac{1}{4}

Keywords: Logarithms

Learn more about Logarithms at:

  • brainly.com/question/11921476
  • brainly.com/question/10633485
  • brainly.com/question/5758530

#learnwithBrainly

4 0
3 years ago
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