Answer:
<h2>
The answer is B. $8.82</h2>
Step-by-step explanation:
given data
mints $0.96 per pound
chocolates $4.70 per pound
lollipops $0.07 each
cost of each items based on the amount item
mints $0.96*0.75= 0.72
chocolates $4.70*1.5=7.05
lollipops $0.07*15=1.05
total cost= 0.72+7.05+1.05
total cost=$8.82
So first remember the order of operations. Parentheses, exponents, multiply, divide, addition, subtraction.
So we have the following problem.
-(3*2)2+4*3-2
Multiply the three and two inside the parentheses first.
-6*2+4*3-2
Next multiply any numbers that need multiplied.
-12+12-2
Next add and subtract from left to right to get an answer of...
-2
Hope this helps!
Tentukan mean, median, dan modus dari 7,4,5,6,7,4,5,7,8,9, dan 6
Anvisha [2.4K]
Answer:
<h3>Median = 6</h3><h3>Mean = 6,1</h3><h3>Modus = 7</h3>
Step-by-step explanation:
Median (Nila tengah setelah data diurutkan)
= 4, 4, 5, 5, 6, 6, 7, 7, 7, 8, 9
= 6
Mean
= jumlah data / banyak data
= 4+4+5+5+6+6+7+7+7+8+9/11
= 68/11
= 6,1
Modus
= data yang paling banyak muncul
= 7 (muncul sebanyak 3 kali)
Answer:
x = -3
y = 0
Step-by-step explanation:
<u>Given</u><u> </u><u>equations</u><u> </u><u>:</u><u>-</u><u> </u>
<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u>
<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u>
<u>From</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u><u> </u><u> </u>
<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u>
<u>-x</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>-</u><u> </u><u>2</u><u>y</u><u> </u>
<u>x</u><u> </u><u>=</u><u> </u><u>2</u><u>y</u><u> </u><u>-</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u>
<u>From</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u><u> </u>
<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u>
<u>2</u><u>x</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>+</u><u> </u><u>3</u><u>y</u><u> </u>
<u>
</u>
<u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>
<u>Equating</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>
<u>x</u><u> </u><u>=</u><u> </u><u>x</u><u> </u>
<u>
</u>
4y - 6 = -6 + 3y
4y - 3y = -6 + 6
y = 0
Putting value of y in ( iii )
x = 2y - 3
x = 2 ( 0 ) - 3
x = -3