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stich3 [128]
3 years ago
9

Can somebody help me plz with number 13,14,15,16,17

Mathematics
1 answer:
yuradex [85]3 years ago
5 0
13. There are 3 feet in a yard. You divide 34 by 3 and get 11 with one left     over. So it should be 11 yards and 1 foot.
14.  16 oz. = 1 lb. If you multiply 16 (oz) x 8 (lbs.) you get 128 ounces.
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The depth of a lake is 15.8m
Leya [2.2K]

a. Jada got the measurement 16m.

b. The measured depth differ 0.2m from the actual depth.

c. There is a 12.66% error in calculation.

Step-by-step explanation:

Given,

The depth of lake = 15.8 m

a. Jada accurately measured the depth of the lake to the nearest meter. What measurement did Jada get

When a digit after the decimal point is 5 or more than that, the number before decimal is rounded to the next number.

Therefore,

15.8 rounded to nearest number is 16.

The depth of lake measured by Jada = 16 m

Jada got the measurement 16m.

b. By how many meters does the measured depth differ from the actual depth?

Difference = Approx - Exact

Difference = 16 - 15.8 = 0.2m

The measured depth differ 0.2m from the actual depth.

c. Express the measurement error ad a percentage of the actual depth.

Percent error = \frac{|approx-exact|}{exact}*100

Percent error = \frac{|2|}{15.8}*100 = \frac{200}{15.8}

Percent error = 12.66%

There is a 12.66% error in calculation.

Keywords: percent, error

Learn more about percent at:

  • brainly.com/question/9323337
  • brainly.com/question/9103248

#LearnwithBrainly

4 0
3 years ago
How many solutions does 5a-3=5a-3
olya-2409 [2.1K]

Answer:

0

Step-by-step explanation:

5a-3=5a-3

or, 5a-5a=-3+3

or, 0=0 answer

4 0
2 years ago
Help me <br> pleeeeease and please do it fast cause i need it quickly
Luden [163]

Answer:

i think its c

Step-by-step explanation:

Hope this helps!!

6 0
3 years ago
Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation o
Ierofanga [76]

Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, \mu = 5.5 mins

Standard deviation, \sigma = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒ z_1=\frac{x_1-\mu}{\sigma}   ...raw score,x_1 = 5.4

⇒ Plugging the values.

⇒ z_1=\frac{5.4-5.5}{0.4}

⇒ z_1=-0.25  

For raw score 5.5 the z score is.

⇒ z_2=\frac{5.8-5.5}{0.4}  

⇒ z_2=0.75

Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒ P(5.4

⇒ P(-0.25

⇒ P(z

⇒ z(1.5)=0.7734 and z(-0.25)=0.4013.<em>..from z -score table.</em>

⇒ 0.7734-0.4013

⇒ 0.3721

To find the percentage we have to multiply with 100.

⇒ 0.3721\times 100

⇒ 37.21 %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

4 0
3 years ago
12 (5x-1) = 7x (2x +2)
murzikaleks [220]
60x - 12 = 14x^2 + 14x

Subtract 60x & Combine

-12 = 14x^2+14x-60x
-12= 14x^2-46x
Add 12
0= 14x^2-46x-14
7 0
3 years ago
Read 2 more answers
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