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Vladimir79 [104]
2 years ago
12

Help? i added the drop downs in order

Mathematics
1 answer:
Flauer [41]2 years ago
5 0

Answer:

The perpendicular bisector theorem states that if XY bisects another line segment AB, creating perpendicular angles, then, the line segment XA & XB are congruent to each other.

The length of WY is 7.3. Since WZ is perpendicular to line XY, XZ= ZY and WX= WY. Since WX= 7.3, then WY also must equal 7.3

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What value of t satisfies the equation t/8-3/4=9/16
Y_Kistochka [10]
The answer to t is t= 21/2
5 0
3 years ago
Read 2 more answers
Standard deviation of 300, 785, 670, 187, 760, 724
White raven [17]
STEP 1

Mean = \frac{300+785+670+187+760+724}{6} = \frac{3426}{6}=571

STEP 2

Subtract the mean from each data value then square the answer
300-571=-271 ⇒ (-271)^{2} =73441
785-571=214 ⇒ (214)^{2}=45796
670-571=99 ⇒ (99)^{2}=9801
187-571=-384 ⇒ (-384)^{2} =147456
760-571=189 ⇒ 189^{2}=35721
724-571=153 ⇒ 153^{2} =23409

STEP 3

Add the squared answers in STEP 2 then find the mean

\frac{73441+45796+9801+147456+35721+23409}{6}= \frac{335624}{6}=55937.3

STEP 4

Square root the answer in STEP 3
\sqrt{55937.3} =236.5

Standard deviation is 236.5
 
7 0
3 years ago
A jeweler wants to make 14 grams of an alloy that is precisely 75% gold.. The jeweler has alloys that are 25% gold, 50% gold, &a
Goryan [66]

Given that the jeweler has alloys that are 25% gold, 50% gold, and 82% gold.

As he wants to make 14 grams of an alloy by adding two different alloys that is precisely 75% gold, so one alloy must have a percentage of gold more than 75%.

One alloy is 82% gold and, the second can be chosen between 25% gold, 50% gold, so there are two cases.

Case 1: 82% gold + 50% gold

Let x grams of 82% gold and y  grams of 50% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 50% of y

\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times y \\\\

\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times (14-x)  [as x+y=14]

\Rightarrow 75 \times 14 = 82 \times x + 50 \times (14-x)  \\\\\Rightarrow 75 \times 14 = 82 \times x + 50 \times14-50\times x \\\\\Rightarrow 75 \times 14 = 32 \times x + 50 \times14 \\\\\Rightarrow 32 \times x =75 \times 14 - 50 \times14 \\\\

\Rightarrow x =(25 \times 14)/32=10.9375 grams

and y = 14-x= 14-10.9375=3.0625 grams.

Hence, 10.9375 grams of 82% gold and 3.0625  grams of 50% gold added to make 14 grams of 75% gold.

Case 2: 82% gold + 25% gold

Let x grams of 82% gold and y  grams of 25% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 25% of y

\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times y \\\\\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times (14-x) \\\\ \Rightarrow 75 \times 14 = 82 \times x + 25 \times (14-x)  \\\\\Rightarrow 75 \times 14 = 82 \times x + 25 \times14-25\times x \\\\\Rightarrow 75 \times 14 = 57 \times x + 25 \times14 \\\\\Rightarrow 57 \times x =75 \times 14 - 25 \times14 \\\\

\Rightarrow x =(50 \times 14)/57=12.28 grams

and y = 14-x= 14-12.28=1.72 grams.

Hence, 12.28 grams of 82% gold and 1.72  grams of 50% gold added to make 14 grams of 75% gold.

3 0
2 years ago
3t>5t+12 please help me? i suck at math
Alex777 [14]
<span>3t>5t+12
Subtract 5t from both sides to get the variable on only one side.
-2t>12
Divide both sides by -2
Final Answer: t< -6</span>
3 0
3 years ago
A server worked a shift where he served 13 entrees, 8 appetizers, and 6 desserts . What percentage of his servings were
ryzh [129]

Answer:

entrees 48% appetizers 30% desserts 22%

Step-by-step explanation:

18+8+6=27

13/27

8/27

6/27

6 0
3 years ago
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