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koban [17]
2 years ago
11

Pleasse help now and show all work thanks so much

Mathematics
2 answers:
statuscvo [17]2 years ago
8 0
X = 11 cos 22
x = 10.20
Anuta_ua [19.1K]2 years ago
5 0
Ok its been a while since ive done this but all ur doing is plugging in values then multiplying.
So itd be-

cos(22)=x/11
 then multiply both sides by 11x cos(22)=x
next is cos(22)=x/11
then 11cos(22)=x
x=10.199 which rounds to 10.2
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Based on the survey, what is the probability that a person chosen at random is a diabetic patient or an eye patient
KonstantinChe [14]

First, you must find the total number of people. Add together 32, 54, 78, 112, and 96 to get 372. Next, add the number of diabetic patients with the number of patients with eye problems (54 + 112 = 166).

Your fraction is now 116/372. Simplify this to get your answer, 29/93 or 0.31.

6 0
2 years ago
Which shows the expression below simplified?
salantis [7]

Answer:

D

Step-by-step explanation:

(7 × 10-7) × 0.06

7 × 0.06 × 10^-7

0.42 × 10^-7

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7 0
3 years ago
....................
pickupchik [31]

Answer:

complimentary x=5

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90degrees - 65degrees= 254degrees

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3 0
2 years ago
Simplify the expression <br><br><img src="https://tex.z-dn.net/?f=6%285%20%2B%20t%29%20-%203%28t%20%2B%203%29" id="TexFormula1"
Pavel [41]
30+6t - 3t-9
combine 6t and -3t and get 3t
30+3t=-9
subtract 30 on both sides
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3 0
2 years ago
Which values are outliers?
SSSSS [86.1K]
Answer:
0.8 Not Outlier
1.1 Not Outlier
10.2 Not Outlier
10.9 Not Outlier 

Solution:
Arranging the numbers in ascending order:
     0.8   1.1   4.9   5.2   5.8   5.9   6.1   6.1   7.4   10.2   10.9
we can see that the median is 5.9.

We can find the first quartile Q1 by getting the median in the lower half of the data 
     0.8   1.1   4.9   5.2   5.8
that is, Q1 = 4.9

We can find the third quartile Q3 by getting the median for the upper half of the data 
     6.1   6.1   7.4   10.2   10.9
that is, Q3 = 7.4

We subtract Q1 from Q3 to find the interquartile range IQR.
     IQR = Q3 - Q1 = 7.4 - 4.9 = 2.5

We can now calculate for the upper and lower limits:
     upper limit = Q3 + 1.5*IQR = 7.4 + (1.5*2.5) = 11.15
     lower limit = Q1 – 1.5*IQR = 0.8 - (1.5*2.5) = -2.95
There is no data point that lies above the upper limit and below the lower limit, therefore, there are no outliers in the data set.
8 0
3 years ago
Read 2 more answers
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