1and1/4 x 1and1/4 equals what

What is the answer?<br><br> [(12 − 5) • 12] ÷ 6

Sever21 [200]

(7 × 12 )/ 6
= 7×2
= 14

6 0

4 years ago

You bike 1 mile the first day of your training, 1.3 miles the second day, 1.9 miles the third day, and 3.1 miles the fourth d

JulsSmile [24]

Answer:

19.9 miles

Step-by-step explanation:

In this problem we have:

$d_1=1 mi$ is the distance travelled during the 1st day

$d_2=1.3 mi$ is the distance travelled during the 2nd day

$d_3=1.9 mi$ is the distance travelled during the 3rd day

$d_4=3.1 mi$ is the distance travelled during the 4th day

We notice that the difference between the distance travelled on the (n+1)-th day and the distance travelled on the n-th day doubles every day. In fact:

$d_2-d_1=0.3\\d_3-d_2=2\cdot 0.3 = 0.6\\d_4-d_3=2\cdot 0.6 = 1.2$

Which can be rewritten using the general formula:

$d_{n+1}-d_n=2(d_n-d_{n-1})$

This means that

$d_{n+1}=d_n+2(d_n-d_{n-1})$

By applying this formula recursively, we can find the 7th term, which is the distance travelled on the 7th day:

$d_1=1\\d_2=1.3\\d_3=1.9\\d_4=3.1\\d_5=3.1+2\cdot 1.2=5.5\\d_6=5.5+2\cdot 2.4=10.3\\d_7=10.3+2\cdot 4.8=19.9 mi$

So, the distance travelled on the 7th day is 19.9 miles.

7 0

4 years ago

PLEASE HELP MEE!!!!!!!!!!!!!!!!!!!

Dmitriy789 [7]

Answer:

a b c

plz give branliest

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

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

So

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

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