All categories

3 years ago

60÷5(7-5) what is the answer to this math equation?

Mathematics

2 answers:

Mumz [18]3 years ago

8 0

For this case we have the following expression:

60 ÷ 5 (7-5)

Applying the distributive property we have:

60 ÷ ((5 * 7) - (5 * 5))

Rewriting we have:

60 ÷ (35-25)

60 ÷ 10

Finally, dividing both numbers we have:

60 ÷ 10 = 6

Answer:

The answer to this math equation is:

60 ÷ 5 (7-5) = 6

Ad libitum [116K]3 years ago

7 0

24 is the answer because you do 7-5 because it is in parentheses then you divide 60÷5 and that leaves 12 × 2 so when there's two numbers left together you multiply 12 x 2

You might be interested in

10. V=720 cm L=? P=15 cm T=8 cm A. 6 cm B. 8 cm C.10 cm D. 12 cm

grin007 [14]

Answer:

6 cm

Step-by-step explanation:

Given the information :

V=720 cm

L=?

P=15 cm

T=8 cm

The volume = L * P * T

V = L * P * T

720 = L * 15 * 8

720 = L * 120

L = 720 / 120

L = 6 cm

5 0

2 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider