Answer: x=3
Step-by-step explanation: Step 1: Simplify both sides of the equation.
21x−5=7+17x
21x+−5=7+17x
21x−5=17x+7
Step 2: Subtract 17x from both sides.
21x−5−17x=17x+7−17x
4x−5=7
Step 3: Add 5 to both sides.
4x−5+5=7+5
4x=12
Step 4: Divide both sides by 4.
= 
So Final Answer: <u>x=3</u>
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12 is 150% of what number?
answer- 8
5 is 50% of what number?
answer- 1000
10%of what number is 300?
answer- 3000
5% of what number is 72?
answer- 1440
20 is 80% of what number?
answer- 25
<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>x^</em><em>2</em><em>-</em><em>1</em><em>.</em>
<em>EXPLANATION</em><em>:</em>
<em>To</em><em> </em><em>be</em><em> </em><em>a</em><em> </em><em>polynomial</em><em>,</em><em> </em><em>the</em><em> </em><em>power</em><em> </em><em>of</em><em> </em><em>each</em><em> </em><em>term</em><em> </em><em>must</em><em> </em><em>be</em><em> </em><em>a</em><em> </em><em>whole</em><em> </em><em>number</em><em>.</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>helpful</em><em> </em><em>to</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
<em>~</em><em>p</em><em>r</em><em>a</em><em>g</em><em>y</em><em>a</em>
Answer:
sample size n would be 149305 large
Value of n (149305) is too high, this will be the practical problem with attempting to find this confidence interval
Step-by-step explanation:
Given that;
standard deviation α = 150 min
confidence interval = 99%
since; p( -2.576 < z < 2.576) = 0.99
so z-value for 99% CI is 2.576
E = 1 minutes
Therefore
n = [(z × α) / E ]²
so we substitute
n = [(2.576 × 150) / 1 ]²
n = [ 386.4 ]²
n = 149304.96 ≈ 149305
Therefore sample size would be 149305 large
Value of n is too high, that would be the practical problem with attempting to find this confidence interval
Answer:
Answer in the picture: x = 1 | y = -5
Step-by-step explanation:
Not sure if that's the answer you're looking for but I think that's it