Arithmetic, geometric, or neither 5, 1, 1, 5...

Mustafaâ€™s soccer team is planning a school dance as a fundraiser. The dj charges $200 and decorations cost $100. The team deci

Kaylis [27]

Question is Incomplete;Complete question is given below.

Mustafa’s soccer team is planning a school dance as a fundraiser. The DJ charges $200 and decorations cost $100. The team decides to charge each student $5.00 to attend the dance. If n represents the number of students attending the dance, which equation can be used to find the number of students needed to make $1,500 in profit?

$5n - 300 = 1,500\\\\5n + 300 = 1,500\\ \\5n - 200 + 100n = 1,500\\\\5n - 100 -200n = 1,500$

Answer:

$5n - 300 = 1,500$

Step-by-step explanation:

Given:

Charges of DJ = $200

Decoration charges = $100

Profit need to be made = $1500

Charges per student to attend the dance = $5

We need to write the equation to find the number of students for to attend the dance in order to make profit.

Solution:

Let the number of student be represented by 'n'.

Now we can say that;

Profit to be made should be equal to Charges per student to attend the dance multiplied by number of students minus sum of Charges of DJ and Decoration charges.

framing in equation form we get;

$5n-(200+100)=1500\\\\5n-300=1500$

Hence the equation to find the number of students attending the dance is $5n - 300 = 1,500$

8 0

2 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

3 years ago

If possible may someone help me please?

Orlov [11]

Answer:

Given, ∠MNP=107 degree.

Therefore, ∠QPN= 107 degree. [as, MN=PQ]

We know, sum of four angles of a quadrilateral is 360 degree.

And as MN=PQ, ∠NMQ=∠MQP.

Therefore, ∠MQP= {360 degree-( 107degree+ 107 degree)}/2

=( 360 degree-214 degree)/2

= 146 degree/2

=73 degree.

hope this will help you.

Step-by-step explanation:

4 0

2 years ago

Please answer I will be giving out brainliest and 2<br>7 points.

maria [59]

1. 21 peanuts per minute
2. 9 texts per minute
3. 28 pages per hour OR half a page per minute

4. The first option is the cheapest. You can find what one ounce costs for each barrel by dividing the cost by the total ounces.

6 0

3 years ago

Which of the following expressions are equivalent? Justify your reasoning.. . a.4√x3. . 1. x−1. . b.10√x5•x4•x2. . c.x. 1. 3. •x

sladkih [1.3K]

A ) $\sqrt[4]{ x^{3} } = x^{3/4}$
B ) $\frac{1}{ x^{-1} } = x$
C ) $\sqrt[10]{ x^{5} * x^{4}* x^{2} } = \sqrt[10]{ x^{11} }= x^{11/10}$
D ) $x^{1/3} * x^{1/3} * x^{1/3} = x$
Therefore: B = D

7 0

3 years ago