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3 years ago

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Ainsley walks 3/5 mile in 1/3 of a day. Sarah walks 5/8 mile in 2/6 of a day. Who walked farther, Ainsley or Sarah? Find the uni

t rate for each girl and then tell who you think walked farther with the correct unit rate given below. *

1 answer:

Nutka1998 [239]3 years ago

4 0

$Ainsley \: walked \: in \: \frac{1}{3} \: \: of \: a \: day \ \: = \frac{3}{5} mile$

$= \frac{1}{3} \times 24 = \frac{3}{5}$

$8 \: hours = \frac{3}{5} \: miles$

$Sarah \: walked \: in \: \frac{2}{6} of \: a \: day \: = \frac{5}{8} miles$

$\frac{2}{6} \times 24 = \frac{5}{8}$

$8 \: hours \: = \frac{5}{8} \: miles$

Let us see who walked more :

$Ainsley \: walked \: = \frac{3 \times 8}{40}$

$= \frac{24}{40} \: miles$

$Sarah \: walked \: = \frac{5 \times 5}{40}$

$= \frac{25}{40} \: miles$

$\frac{24}{40} < \frac{25}{40}$

Sarah walked farther than Ainsley .

$\frac{25}{40} - \frac{24}{40} = \frac{1}{40} \: miles$

∴ Sarah walked farther than Ainsley by 1/40 miles .

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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

Help pls don't answer if you don't know also I’ll give brainliest

il63 [147K]

Evaluate for x=4,y=4,z=4

|(2)(4)−4+(3)(4)|

|(2)(4)−4+(3)(4)|

=16

Answer=16

(PLEASE GIVE BRAINLIEST)

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3 years ago

Julia and Cody are working together on solving math problems. For every one problem that Julia

slega [8]

Answer:

Cody has solved (12 × 12) = 144 problems.

Step-by-step explanation:

For every one problem that Julia completes, Cody completes twelve.

If Julia Completes x problems and Cody completes y problems, then we can write y = 12x ........ (1)

Now, given that the number of problems solved by Cody is one hundred twenty more than two times the number of problems solved by Julia.

Hence, 2x + 120 = y ......... (2)

Now, from equations (1) and (2) we get,

2x + 120 = 12x

⇒ 10x = 120

⇒ x = 12

Therefore, Cody has solved (12 × 12) = 144 problems. (Answer)

4 0

3 years ago

last month , he sold 50 chickens and 30 ducks for $550 . This month , he sold 44 chickens and 36 ducks for $532. How much does a

Kazeer [188]

Hello there, my fellow human being!
So, the chicken costs $8 and the duck costs $5, here's why.

Let's say x is the cost of a chicken while y is the cost of a duck.
We can make two linear equations using the information above.
Last month, he sold 50 chickens and 30 ducks for $550: 50x + 30y= 550
This month, he sold 44 chicken and 36 ducks for $532: 44x + 36y = 532

50x/10 + 30y/10 = 440/10
44x/4 + 36y/4 = 532/4

5x + 3y = 44
11x + 9y = 133

So, now that we have our answer simplified, we have to use elimination to solve this system of equations. But, first we need to make sure that at least one of our variables is able to be canceled out.
Let's multiply this equation by -3.
(5x + 3y = 44) * -3.

-15x-9y=-165.

11x + 9y = 133
-15x - 9y = -165
______________
-4x/4 = -32/4
x = 8
11x + 9y = 133
11(8) + 9y = 133
88 + 9y = 133
-88 -88
________________
9y/9 = 45/9
y = 5.

4 0

3 years ago

Read 2 more answers

2. Kelsea needs a test average of at least 90 to get an "A-" this marking period in math. Her three test grades

Luda [366]

Answer:

<u>Equation: 87 + 91 + 86 + x = 360</u>

<u>Solution: 264 + x = 360</u>

<u>x = 360 - 264</u>

<u>x = 96</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

First three Kelsea's test grades : 87, 91 and 86

2. What score must she get on her fourth test to receive at least an A- ?

Define variable: x that represents the grade needed by Kelsea on her fourth test to receive at least an A-

Equation: 87 + 91 + 86 + x = 360

Solution: 264 + x = 360

x = 360 - 264

x = 96

<u>Now you can understand if the previous work you did is correct</u>

6 0

4 years ago