Write the sum using summation notation <br><br><br> 729 + 1000 + 1331 + 1728 + ... + n^3

Logan made a treasure hunt map. If each 1 in on the scale drawing equals 5 ft, then what is the scale factor? What is an equatio

emmasim [6.3K]

Answer:

1=12=13

Step-by-step explanation:

8 0

3 years ago

Samantha sends her son, Barry, to a preschool center on certain days. The cost of preschool is $45 per day along with a fixed mo

MAXImum [283]

Answer:

Step-by-step explanation:

We are given fixed monthly charge = $70.

The cost of preschool per day = $45.

Number of days = d.

Total cost of d days = cost per day × number of days + fixed monthly charge.

Therefore, we get equation

880 = 45×d+70

Now, we need to solve the equation for d.

Subtracting 70 from both sides, we get

880-70 = 45d +70-70

810=45d

Dividing both sides by 45, we get

$\frac{810}{45} =\frac{45d}{45}$

18=d.

Therefore,<em> 18 days Barry attended preschool last month.</em>

<em>Therefore, correct option is D option.</em>

4 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

4 years ago

2/9+5/6 in simplest form

never [62]

2/9+5/6 in simplest form is 1 1/18
Hope this helped :P

6 0

3 years ago

Read 2 more answers

Use the graphs below to help you answer the question. Which of the following is the best approximation to a solution of the equa

brilliants [131]

Answer:

2 is the correct option.

Step-by-step explanation:

We have the equation $e^x=4x+1$ .

We will plot the graphs of the function $e^x$ and $4x+1$ .

Then, the intersection points of both the graph will be the solution of the equation $e^x=4x+1$ .

From the graphs, we see that,

The intersection points of the graphs are (0,1) and (2.337,10.347).

The solution of the equation is the value of 'x' co-ordinate where they intersect.

So, we get, x = 2 is the correct option.

4 0

3 years ago

Write the sum using summation notation 729 + 1000 + 1331 + 1728 + ... + n^3