Answer:
4. < ACB = 90° and < CBA = 52°
5. <SPQ+ <SRQ = 180°
so, <SPQ= < QRT = 100°
You can set them equal to each other so -3x+4=4x-10 and then you add 3x and 10 on both sides and get7x=14 and then divided both sides by 7 and get x = 2 and check by plugging in and you get -2 for y on both so solution is x=2
Answer:
−2(−9x+5)−4x=2(x−5)−4
Step 1: Simplify both sides of the equation.
−2(−9x+5)−4x=2(x−5)−4
(−2)(−9x)+(−2)(5)+−4x=(2)(x)+(2)(−5)+−4(Distribute)
18x+−10+−4x=2x+−10+−4
(18x+−4x)+(−10)=(2x)+(−10+−4)(Combine Like Terms)
14x+−10=2x+−14
14x−10=2x−14
Step 2: Subtract 2x from both sides.
14x−10−2x=2x−14−2x
12x−10=−14
Step 3: Add 10 to both sides.
12x−10+10=−14+10
12x=−4
Step 4: Divide both sides by 12.
12x
12
=
−4
12
x= -1/3
Step-by-step explanation:
Good luck! Hope this helps! <3
Answer:
63°
Step-by-step explanation:
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hope it helps...
have a great day!!
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The median number of minutes for Jake and Sarah are equal, but the mean numbers are different.
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For this, you never said the choices, but I’ve done this before, so I’m going to use the answer choices I had, and hopefully they are right.
Our choices are -
• The median number of minutes for Jake is higher than the median number of minutes for Sarah.
• The mean number of minutes for Sarah is higher than the mean number of minutes for Jake.
• The mean number of minutes for Jake and Sarah are equal, but the median number of minutes are different.
• The median number of minutes for Jake and Sarah are equal, but the mean number of minutes are different.
————————
So to answer the question, we neee to find the median and mean for each data set, so -
Jack = [90 median] [89.6 mean]
Sarah = [90 median] [89.5 mean]
We can clearly see the median for both is 90, so we can eliminate all the choices that say they are unequal.
We can also see that Jack has a higher mean (89.6) compared to Sarah (89.5).
We can eliminate all the choices that don’t imply that too.
That leaves us with -
• The median number of minutes for Jake and Sarah are equal, but the mean number of minutes are different.
