All categories

3 years ago

Evaluate. 16 • 3 + [(17 – 3) ÷ 2] – 43 What is the answer

Mathematics

2 answers:

natita [175]3 years ago

8 0

16 × 3 + [(17 – 3) ÷ 2] – 43
16 × 3 + [(14) ÷ 2] – 43
16 × 3 + 7 – 43
48 + 7 – 43
55 – 43
12

scoundrel [369]3 years ago

6 0

$16\cdot3+[(17-3)\div2]-43=$
$= 48+(14\div2)-43$
$=48+7-43$
$=55-43$
$=\boxed{\boxed{12}}$

You might be interested in

Find |37|. 0 -37 37 i have 3 minutes to finish HELPP :(

valentina_108 [34]

It’s -37! | this means the opposite of the number ;)

6 0

3 years ago

Ali spends 30% monthly income on food articles, 40% of the remaining on conveyance and clothes and saves 50% of the remaining. I

Romashka [77]

Ali saves Rs 2576 every month

Step-by-step explanation:

Step 1: Given Ali's monthly salary = Rs 18400. Calculate the amount spent on food articles

Amount spent on food = 30% of 18400 = 30/100 × 18400 = Rs 5520

Step 2: Find the remaining amount.

Remaining amount = 18400 - 5520 = Rs 12880

Step 3: Find amount spend on conveyance and clothes

Amount spent on conveyance and clothes = 40% of 12880

= 40/100 × 12880 = Rs 5152

Step 4: Find his monthly savings.

Monthly savings = 50% of 5152 = 50/100 × 5152 = Rs 2576

5 0

3 years ago

Read 2 more answers

16. Find the final balance in an account with $730 invested at 3% annual simple interest for 4 years. (1 point) $735.48 $87.60 $

podryga [215]

Answer:

$817.60

Step-by-step explanation:

Simple interest has the formula:

Simple Interest = Principal * rate * time

here,

Principal amount is $730

The rate of interest is 3%, which is 3/100 = 0.03

The time length in years is given as 4

We find the interest amount first:

SI = P * r * t

SI = 730 * 0.03 * 4

SI = 87.6

We are asked to find FINAL BALANCE, which is PRINCIPAL + INTEREST

So,

Final Balance = 730 + 87.6 = $817.60

7 0

3 years ago

What is 5/10 = x/100

Monica [59]

Answer:

= 50

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Geometric Series assistance

Levart [38]

we have been asked to find the sum of the series

$\sum _{n=1}^5\left(\frac{1}{3}\right)^{n-1}$

As we know that a geometric series has a constant ratio "r" and it is defined as

$r=\frac{a_{n+1}}{a_n}=\frac{\left(\frac{1}{3}\right)^{\left(n+1\right)-1}}{\left(\frac{1}{3}\right)^{n-1}}=\frac{1}{3}$

The first term of the series is $a_1=\left(\frac{1}{3}\right)^{1-1}=1$

Geometric series sum formula is

$S_n=a_1\frac{1-r^n}{1-r}$

Plugin the values we get

$S_5=1\cdot \frac{1-\left(\frac{1}{3}\right)^5}{1-\frac{1}{3}}$

On simplification we get

$S_5=\frac{121}{81}$

Hence the sum of the given series is $\frac{121}{81}$

5 0

3 years ago