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sertanlavr [38]
3 years ago
8

18.5−1.73 what is this??

Mathematics
1 answer:
Debora [2.8K]3 years ago
5 0
The answer for this is 16.77
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ILL GIVE BRAINLIEST TO THE RIGHT ANSWER!!!!!
Paraphin [41]

Answer:

-3<lxl<5

Step-by-step explanation:

take least number than use x in absolute value bars than take highest number sorry if this is wrong I'm just a kid

4 0
3 years ago
A swimming pool had 2.5 liters of water in it. Some water evaporated, and then the pool only had 2 million liters of water in it
svet-max [94.6K]
Water in the swimming pool initially = 2.5 million liters 
Water in the swimming pool after evaporation = 2 million liters 
Therefore, volume of water evaporated = 2.5 million liters - 2 million liters 
 =0.5 million liters 
To find the percent of water evaporated, divide the volume of water evaporated by the volume of water in the swimming pool initially and then multiply it by 100. 
 Percent of water evaporated =( 0.5 / 2.5) x 100  =20%
6 0
3 years ago
Wei has 150.00 to make a garland using 60-cent balloons
andreev551 [17]

Answer:

The answer is 0.60b + 0.30w = 150; 60 + 0.30w = 150

Step-by-step explanation:

7 0
3 years ago
The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 55 o
Usimov [2.4K]

Answer:

a) For this case and using the empirical rule we can find the limits in order to have 9% of the values:

\mu -2\sigma = 55 -2*6 =43

\mu +2\sigma = 55 +2*6 =67

95% of the widget weights lie between 43 and 67

b) For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

100 -0.15-2.5 = 97.85

c) We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%

Step-by-step explanation:

For this case our random variable of interest for the weights is bell shaped and we know the following parameters.

\mu = 55, \sigma =6

We can see the illustration of the curve in the figure attached. We need to remember that from the empirical rule we have 68% of the values within one deviation from the mean, 95% of the data within 2 deviations and 99.7% of the values within 3 deviations from the mean.

Part a

For this case and using the empirical rule we can find the limits in order to have 9% of the values:

\mu -2\sigma = 55 -2*6 =43

\mu +2\sigma = 55 +2*6 =67

95% of the widget weights lie between 43 and 67

Part b

For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

100 -0.15-2.5 = 97.85

Part c

We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%

4 0
4 years ago
A local school has was evaluated the the Education Department and received the
Angelina_Jolie [31]

Answer: 79

Step-by-step explanation:

Given the following :

Graduation Rates:

Grade:82

Weight: 50%

Passing Rates:

Grade: 75

Weight: 30%

Enrollment Numbers:

Grade: 60

Weight: 10%

Career readiness:

Grade: 95

Weight: 10%

Overall score : Σ(grade × weight)

(82 × 50%) + (75 × 30%) + (60 × 10%) + (95 × 10%)

= (41 + 22.5 + 6 + 9.5)

= 79

Hence, overall score is 79

5 0
3 years ago
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