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4 years ago

If a donut costs 50 cents and a cookie costs 35 cents, write an

Mathematics

1 answer:

zhannawk [14.2K]4 years ago

5 0

Answer:

50d+35c= total cost

Step-by-step explanation:

50 cents for a donut, so 50 * d = total cost for donuts

35 cents for a cookie, so 35*c = total cost for cookies

add them up for the answer!

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HI I NEED HELP IVE BEEN DOING HOMEWORK FOR 6 HOURS ummmmm anyways, how do i express this (y^3•y^6 - in picture) in exponential f

LUCKY_DIMON [66]

y^3 = y * y * y

y^6 = y * y * y * y * y * y

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3 years ago

Read 2 more answers

What is the maturity value of $1 at a compound interest rate of 9 percent over 4 periods

Vaselesa [24]

Your answer is 1. Hoped this has helped

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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

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

WILL GIVE BRAINLIEST I ONLY HAVE 20 MINS

tamaranim1 [39]

Answer:

The answer is (D)

Adding the 5 and 3 then multiplying them by the numbers in the brackets in order to remove them from the brackets

In other words Simplifying the equation

8 0

2 years ago

A hospital recorded the weights, in ounces, of newborn babies for two weeks. The

Gwar [14]

Answer:

The standard deviation for week two was about 3 ounces more than the standard deviation for week one

Step-by-step explanation:

Given

$Week\ 1: 128, 105, 80, 82, 96, 98, 87, 100, 112, 126$

$Week\ 2: 75, 85, 90, 97, 89, 105, 110, 127, 129, 130$

See attachment for options

Required

The true statement

Checking the standard deviation

For week 1

Calculate the mean:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{128+ 105+ 80+ 82+ 96+ 98+ 87+ 100+ 112+ 126}{10}$

$\bar x = \frac{1014}{10}$

$\bar x_1 = 101.4$

Then standard deviation

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

$\sigma_1 = \sqrt{\frac{(128 - 101.4)^2 +............+ (126- 101.4)^2}{10}}$

$\sigma_1 = \sqrt{\frac{2522.4}{10}}$

$\sigma_1 = \sqrt{252.24$

$\sigma_1 = 15.88$

For week 2, we have:

$\bar x = \frac{75+ 85+ 90+ 97+ 89+ 105+ 110+ 127+ 129+ 130}{10}$

$\bar x = \frac{1037}{10}$

$\bar x_2 = 103.7$

Then standard deviation

$\sigma_2 = \sqrt{\frac{(75 - 103.7)^2 +................+ (130- 103.7)^2}{10}}$

$\sigma_2 = \sqrt{\frac{3538.1}{10}}$

$\sigma_2 = \sqrt{353.81$

$\sigma_2 = 18.81$

Compare the standard deviations

$\sigma_1 = 15.88$

$\sigma_2 = 18.81$

Calculate the difference:

$d = \sigma_2 - \sigma_1$

$d = 18.81 - 15.88$

$d = 2.93$

$d \approx 3$

This implies that option (b) is true

4 0

3 years ago