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Elena-2011 [213]
2 years ago
10

Can I have help on this?​

Mathematics
1 answer:
AysviL [449]2 years ago
6 0

Answer:

25°

Step-by-step explanation:

51+x+14=90°

x+65=90°

x=90-65

x=25°

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Two shopping carts are pushed with the same force. One cart has a larger mass than the other. How does the mass of the shopping
ANEK [815]
The correct option would be B. "The cart with the larger mass will accelerate faster."
6 0
3 years ago
Read 2 more answers
Identify the fraction as proper, improper, or mixed . (If the fraction is equivalent to 1, choose "one" in the blank.)
aivan3 [116]
Proper fractions are those whose numerator is less than the denominator.

These are examples of proper fractions:

1/10, 1/9, 1/4, 1/2, 7/8, 221/225, 100/2000.

Improper fractions are those whos numerator is greater than the denominator.

These are examples of improper fractions:

10/3, 9/5, 8/7, 100/99, 5/4, 125/35.

Mixed fractions are made up of one whole number and a fraction (normallly a proper fraction).

These are examples of mixed fractions:

1 1/4 (which is read one and one quarter)

3 2/3 (which is read three and 2 thirds)

10 1/5 (which is read ten and one fith)

3 1/8 (which is read 3 and 1 eigth)

Fractions equivalent to one are those whose numerator is equal to the denominator.

With that information you have the tools to identify your fractions, given that you did not attached them.
4 0
3 years ago
A bulb can either be on or off. A board contains connected to a randomization circuit that lights up a random sequence every tim
kari74 [83]

The question is incomplete. The complete question is :

A bulb can either be on or off. A board contains 20 bulbs connected to a randomization circuit that lights up a random sequence every time it is turned on. What is the probability that all the lights will switch on ?

Solution :

It is given that :

There is board connected to = 20 bulbs

And one can either be put ON or put OFF.

Therefore the chance that a bulb is ON = $\frac{1}{2}$

It is given that every time a bulb turns ON, the board lights up in a random sequence.

Therefore, the probability of all the lights to be switched ON is given by :

$=\left(\frac{1}{2}\right)^{20}$

$=\frac{(1)^{20}}{(2)^{20}}$

$=\frac{1}{1048576}$

7 0
3 years ago
From 11 positive integer scores on a 10-point quiz, the mean is exactly 8, the median is exactly 8, and the mode is exactly 7. f
Zanzabum
<span>3 perfect scores of 10 There can't be 5x10 because the mode (the commonest score) is 7. Hence if there are 5x10 there must be 6 x7, but the median (the middle score) is 8 and this would make at least 12 scores which is impossible. There can't be 4x10. If there were 4 x10 there would be 5x7 (7=mode), 1x8(median) + either another 8 or 9. Assuming the best case scenario (2 x8). The total of the 11 scores = 91. They should add up to 88 (mean = 8, (11x8=88) If there are 3x10, there must be at least 4x7(mode =7), 3more 8+(median = 8), and 1 x less than 7. 3x10, 3x8, 4x7 and 1x6 fits the bill. Mean= (30+24+28+6)/11 = 88/11 = 8 OK Median is the sixth number in the series = 8 OK The mode is the commonest number. Mode = 7 OK</span>
7 0
3 years ago
How do I solve this ?
igomit [66]

Answer:

16

Step-by-step explanation:

because it's g of 7 use the third equation.

(7+1)(7-5) = 8x2 = 16

3 0
2 years ago
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