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3 years ago

What do it equal 1 1/3+ 1 3/7

2 answers:

7 0

1 + 1 = 2

1/3 3/7 can be made equilavent by multiplying

1/3 equals 7/21
3/7 equals 9/21

Added equals 16/21

So the answer is 2 16/21

katen-ka-za [31]3 years ago

6 0

If you would like to calculate 1 1/3 + 1 3/7, you can do this using the following steps:

1 1/3 + 1 3/7 = 4/3 + 10/7 = 28/21 + 30/21 = (28 + 30) / 21 = 58/21 = 2 16/21

The correct result would be 2 16/21.

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Timothy purchased a computer for $1,000.

scoundrel [369]

Answer:

This situation represents [a.) exponential decay

The rate of growth or decay, r, is equal to c.) 0.2

So the value of the computer each year is g.) 80% of the value in the previous year.

It will take [i.) 3 years for the value of the computer to reach $512.

Step-by-step explanation:

Since the value depreciates each year, it represents exponential decay.

The rate of growth or decay, r, is equal to

Depreciates 20%, so r = 0.2.

So the value of the computer each year is ... of the value in the previous year,

Depreciates 20%, so it is 100% - 20% = 80% of the value in the previous year.

It will take x years for the value of the computer to reach $512.

The value of the computer after x years is given by:

$V(x) = V(0)(1-r)^x$

In which V(0) is the initial value and r is the decay rate. So

$V(x) = 1000(1-0.2)^x$

$V(x) = 1000(0.8)^x$

We want to find x for which V(x) = 512. So

$V(x) = 1000(0.8)^x$

$512 = 1000(0.8)^x$

$(0.8)^x = \frac{512}{1000}$

$(0.8)^x = (0.512)$

Applying log to both sides:

$\log{(0.8)^x} = \log{(0.512)}$

$x\log{0.8} = \log{(0.512)}$

$x = \frac{\log{0.512}}{\log{0.8}}$

$x = 3$

So it will take 3 years.

6 0

3 years ago

A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se

bearhunter [10]

Answer:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

Step-by-step explanation:

We know the following info from the problem

$\bar X_1 = 14.5$ sample mean for the group 1

$s_1 = 3.98$ the standard deviation for the group 1

$n_1= 20$ the sample size for group 1

$\bar X_2 = 13.9$ sample mean for the group 2

$s_1 = 4.03$ the standard deviation for the group 2

$n_2= 17$ the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given :

$df = n_1 +n_2- 2 = 20+17-2=35$

The confidence level is 0.9 or 90% and the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 = 0.05$ and the critical value for this case is:

$t_{\alpha/2} = 1.69$

And replacing the info given we got:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

5 0

3 years ago

7-14m=2m=5 please help

Firdavs [7]

Answer:

M=1/6

Step-by-step explanation:

6 0

3 years ago

Let AB be the directed line segment beginning at point A(1,1) and ending at point B(15,-3). Find the point P on the line segment

Svetllana [295]

Answer: it is 3 bc dogs are cancerous, resulting i n pythagoras himself

Step-by-step explanation:

3m/sqrt56 + 9(32x) - t = BOOMER BOOMER BOOMER BOOOMER BOOMERR

OAOAOOAAOOAOAOOAOAOAOOAOAOOAAOOAOAOA

3 0

3 years ago

Mr. Cipalla bought 5 pounds of clay for an art project. He paid $35.35 for clay. What is the unit price per pound of the clay he

Marrrta [24]

Answer:

$7.07

Step-by-step explanation:

35.35/5=7.07

3 0

2 years ago