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3 years ago

Whoever answers will get 60 points. Pls answer based off of the answer choices.

Mathematics

1 answer:

AnnyKZ [126]3 years ago

7 0

Hello from MrBillDoesMath!

Answer:

Discussion:

$350 at 6% = => as "percent" means per 100

350 * (6/100) = => as 350 * 6 = 2100

2100/100 =

Thank you,

MrB

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A fruit company delivers its fruit in two types of boxes: large and small. a delivery of 5 large boxes and 3 small boxes has a t

Hoochie [10]

Let's let the weight of a large box be L, and the weight of a small box be S.

We know that 5 large boxes and 3 small boxes is 120kg, so:
5L + 3S = 120

We also know that 7 large boxes and 9 small boxes is 234kg, so:
7L + 9S = 234

You can multiply the first equation by 3 to get:
15L + 9S = 360

See how now both equations have 9S? We can now subtract one from the other:
(15L+9S) - (7L+9S) = 360-234
8L = 126
L = 15.75

Now sub this value back into an equation:
(5x15.75) + 3S = 120
3S = 41.25
S = 13.75

Double check these values
(7x15.75) + (9x13.75)
= 110.25 + 123.75
=234, which is consistent with above.

So a large box is 15.75kg, and a small box is 13.75kg.

Hope this helped

5 0

3 years ago

When Barbara started her trip, the odometer read 13,487.8. When she ended the trip, it read

solniwko [45]

Who was the 1 president

8 0

3 years ago

Read 2 more answers

Round 38.49 to the nearest whole number

AnnyKZ [126]

38. The nearest whole number to 38.49 is 38 because at .5 you begin rounding up to the nearest whole number.

8 0

3 years ago

In a simple regression analysis for a given data set, if the null hypothesis β = 0 is rejected, then the null hypothesis ρ = 0 i

kati45 [8]

Answer:

Null hypothesis: $\rho =0$

Alternative hypothesis: $\rho \neq 0$

The statistic to check the hypothesis is given by:

$t=\frac{r \sqrt{n-2}}{\sqrt{1-r^2}}$

And is distributed with n-2 degrees of freedom

And the statistic to check the significance of a coeffcient in a regression is given by:

$t_1 = \frac{\hat{\beta_1} -0}{S.E (\hat{\beta_1})}$

For this case is importantto remember that t1 and p value for test of slope coefficient is the same test statistic and p value for the correlation test so then the answer would be:

Always

Step-by-step explanation:

In order to test the hypothesis if the correlation coefficient it's significant we have the following hypothesis:

Null hypothesis: $\rho =0$

Alternative hypothesis: $\rho \neq 0$

The statistic to check the hypothesis is given by:

$t=\frac{r \sqrt{n-2}}{\sqrt{1-r^2}}$

And is distributed with n-2 degrees of freedom

And the statistic to check the significance of a coeffcient in a regression is given by:

$t_1 = \frac{\hat{\beta_1} -0}{S.E (\hat{\beta_1})}$

For this case is importantto remember that t1 and p value for test of slope coefficient is the same test statistic and p value for the correlation test so then the answer would be:

Always

4 0

3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago