3 1/3+(-2 1/4) + 1 5/6

20 POINTS! HELP PLEASE!

nevsk [136]

Answer:

False

Step-by-step explanation:

i did this question and got it right

5 0

3 years ago

Read 2 more answers

On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove

gogolik [260]

Velocity, distance and time:

This question is solved using the following formula:

$v = \frac{d}{t}$

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time $t_1$ , distance of $d_1$ , v = 40. So

$v = \frac{d}{t}$

$40 = \frac{d_1}{t_1}$

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is $v = 60$ .

On the first day, he drove 2 more hours, which means that for the second day, the time is: $t_1 - 2$

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: $d_1 - 20$

Thus

$v = \frac{d}{t}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

System of equations:

Now, from the two equations, a system of equations can be built. So

$40 = \frac{d_1}{t_1}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

Find the total distance traveled in the two days:

We solve the system of equation for $d_1$ , which gets the distance on the first day. The distance on the second day is $d_2 = d_1 - 20$ , and the total distance is:

$T = d_1 + d_2 = d_1 + d_1 - 20 = 2d_1 - 20$

From the first equation:

$d_1 = 40t_1$

$t_1 = \frac{d_1}{40}$

Replacing in the second equation:

$60 = \frac{d_1 - 20}{t_1 - 2}$

$d_1 - 20 = 60t_1 - 120$

$d_1 - 20 = 60\frac{d_1}{40} - 120$

$d_1 = \frac{3d_1}{2} - 100$

$d_1 - \frac{3d_1}{2} = -100$

$-\frac{d_1}{2} = -100$

$\frac{d_1}{2} = 100$

$d_1 = 200$

Thus, the total distance is:

$T = 2d_1 - 20 = 2(200) - 20 = 400 - 20 = 380$

The total distance traveled in two days was of 380 miles.

For the relation between velocity, distance and time, you can take a look here: brainly.com/question/14307500

3 0

3 years ago

What is the product of 1/2 x - 1/4 and 5x^2 - 2x + 6? Write your answer in standard form?

marta [7]

Answer:

for a. you will first solve 5x²-2x+6 with

x=-b ± ✓b²-4ac

----------------

2a

then put the two answers you will get into a bracket then multiply it with 1/2x - 1\4

then write in form of dis :- A ×10^n

Step-by-step explanation:

b. the two products are not the same

3 0

2 years ago

Lin sold 4 more shirts than Greg. Fran sold 3 times as many shirts as Lin. In total the three sold 51 shirts. Which represents t

Alex17521 [72]

<span>Your Answer: D

Why....

You can make equations out of the information
Let L be Lin, G be Greg, and F be Fran

L = 4 + G ---"Lin sold 4 more shirts than Greg"
F = 3L ---"Fran sold 3 times as many shirts as Lin"
F + G + L = 51 ---"In total the three sold 51 shirts"

Use F + G + L = 51
Substitute the equations in for F and L (because you need to know the G) like this....

(3L) + G + (4+G) = 51

You still have a variable besides G in there... you can use the L= 4+G and substitute again so that there are only G's

( 3(4+G) ) + G + (4+G) = 51 ---- SIMPLIFY :D
( 12 + 3G ) + G + 4 + G = 51
12 + 3G + G + 4 + G = 51 ---Combine like terms
16 + 5G = 51 </span>

6 0

3 years ago

An insurance company selected a random sample of 500 clients under 18 years of age and found that 180 of them had had an acciden

Butoxors [25]

Answer:

a) The pooled proportion is p=0.3.

b) P-value = 0.000078

c) Lower bound = 0.0556

d) Upper bound = 0.1644

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the accident proportions differ between the two age groups .

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=500 has a proportion of p1=0.36.

$p_1=X_1/n_1=180/500=0.36$

The sample 2, of size n2=600 has a proportion of p2=0.25.

$p_1=X_1/n_1=150/600=0.25$ .

The difference between proportions is (p1-p2)=0.11.

$p_d=p_1-p_2=0.36-0.25=0.11$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{180+150}{500+600}=\dfrac{330}{1100}=0.3$

The standard error for the difference between proportions can now be calculated as:

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.3*0.7}{500}+\dfrac{0.3*0.7}{600}}\\\\\\s_{p1-p2}=\sqrt{0.00042+0.00035}=\sqrt{0.00077}=0.0277$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.11-0}{0.0277}=\dfrac{0.11}{0.0277}=3.964$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(t>3.964)=0.000078$

As the P-value (0.000078) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the accident proportions differ between the two age groups.

If we want to calculate the bounds of a 95% confidence interval, we start by calculating the margin of error.

For a 95% CI, the critical value for z is z=1.96.

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.96\cdot 0.0277=0.0544$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.11-0.0544=0.05561\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.11+0.0544=0.16439$

The 95% confidence interval for the population mean is (0.0556, 0.1644).

5 0

3 years ago