Rolls are being prepared to go to grocery stores. Divide 72 rolls into 2 groups so the ratio is 3 to 5. (please include your wor

zepelin [54]

Answer: {(x + 2), (x - 1), (x - 3)}

Step-by-step explanation:

Presented symbolically, we have:

x^3 - 2x^2 - 5x + 6

Synthetic division is very useful for determining roots of polynomials. Once we have roots, we can easily write the corresponding factors.

Write out possible factors of 6: {±1, ±2, ±3, ±6}

Let's determine whether or not -2 is a root. Set up synthetic division as follows:

-2 / 1 -2 -5 6

-2 8 -6

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

1 -4 3 0

since the remainder is zero, we know for sure that -2 is a root and (x + 2) is a factor of the given polynomial. The coefficients of the product of the remaining two factors are {1, -4, 3}. This trinomial factors easily into {(x -1), (x - 3)}.

Thus, the three factors of the given polynomial are {(x + 2), (x - 1), (x - 3)}

7 0

3 years ago

Sandra decided to bake blackberry pies to sell at the county fair. She had $180 in her savings

Hatshy [7]

Answer:

275

Step-by-step explanation:

180-130=50

25*9=225

225+50 = 275

6 0

2 years ago

The tennis team is having a fundraiser for uniforms for it’s nine players. the total cost for the uniforms is $759.84 and the te

Sonja [21]

Answer:

C. 9x + 224.36 ≥ 759.84

Step-by-step explanation:

If each team member raises x dollars, the 9 team members will have raised 9x dollars. That amount added to the amount they already have must equal or exceed the amount required:

9x + 224.36 ≥ 759.84 . . . . matches selection C

5 0

3 years ago

A company wants to find out if the average response time to a request differs across its two servers. Say µ1 is the true mean/ex

lorasvet [3.4K]

Answer:

a) The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

Test statistic t=0.88

The P-value is obtained from a t-table, taking into acount the degrees of freedom (419) and the type of test (two-tailed).

b) A P-value close to 1 means that a sample result have a high probability to be obtained due to chance, given that the null hypothesis is true. It means that there is little evidence in favor of the alternative hypothesis.

c) The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

d) The consequences of the confidence interval containing 0 means that the hypothesis that there is no difference between the response time (d=0) is not a unprobable value for the true difference.

This relate to the previous conclusion as there is not enough evidence to support that there is significant difference between the response time, as the hypothesis that there is no difference is not an unusual value for the true difference.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that there is significant difference in the time response for the two servers.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

$M_d=M_1-M_2=12.5-12.2=0.3$

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

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{3^2}{196}+\dfrac{4^2}{225}}\\\\\\s_{M_d}=\sqrt{0.046+0.071}=\sqrt{0.117}=0.342$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.3-0}{0.342}=\dfrac{0.3}{0.342}=0.88$

The degrees of freedom for this test are:

$df=n_1+n_2-1=196+225-2=419$

This test is a two-tailed test, with 419 degrees of freedom and t=0.88, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>0.88)=0.381$

As the P-value (0.381) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that there is significant difference in the time response for the two servers.

<u>Confidence interval </u>

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

The estimated standard error of the difference is s_Md=0.342.

The critical t-value for a 95% confidence interval and 419 degrees of freedom is t=1.966.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=1.966 \cdot 0.342=0.672$

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

$LL=M_d-t \cdot s_{M_d} = 0.3-0.672=-0.372\\\\UL=M_d+t \cdot s_{M_d} = 0.3+0.672=0.972$

The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

7 0

3 years ago

Laura sees a horse pulling a buggy. She wonders how it can accelerate if the action of the horse pulling the cart would causes a

KatRina [158]

Answer:

The net forces exerted on the horse and cart are not the same, so they are not balanced forces.

Step-by-step explanation:

Please see the Newton's 2nd Law which states that an object accelerates if there is a net or unbalanced force on it. In this scenario there is just one force exerted on the wagon i.e: the force that the horse exerts on it. The wagon accelerates because the horse pulls on it. And the amount of acceleration equals the net force on the wagon divided by its mass.

As there are two forces the push and pull the horse; the wagon pulls the horse backwards, and the ground pushes the horse forward. The net force is determined by the relative sizes of these two forces.

If the ground pushes harder on the horse than the wagon pulls, there is a net force in the forward direction, and the horse accelerates forward, and if the wagon pulls harder on the horse than the ground pushes, there is a net force in the backward direction, and the horse accelerates backward.

If the force that the wagon exerts on the horse is the same size as the force that the ground exerts, the net force on the horse is zero, and the horse does not accelerate.

7 0

3 years ago

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