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aleksandr82 [10.1K]
2 years ago
8

Store A advertises a sale for 3 loves of bread for $5.94. Explain how to use an equivalent rate to find the unit price of the br

ead at store A.
Mathematics
2 answers:
tatyana61 [14]2 years ago
5 0

Hello,

To find the Unit Price of the loaves of bread just divide the cost by the amount of loaves it costed.

5.94 / 3 = $1.98

Each loaf (Unit Price) costs $1.98!

Rainbow [258]2 years ago
4 0

Because 3 loaves of bread costs $5.94, to find the unit price, you just divide the price by the amount of loaves.

$5.94÷3≈$1.98

The unit price is $1.98 per loaf of bread.

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Which expression shows the distance on the number line between 2 and −7?
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D because they are 9 ponits between them so the abs are going going to remove the negative
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Danielle had planned her gift basket to be priced at $50 with a 40% gross margin. She forgot packaging costs which will be $3 pe
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B 34% would be the gross margin
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An inspector checks 98 cell phones and finds that 2 of them are defective. A company has 850 of these phones. How many of the co
jekas [21]

Answer:

17

Step-by-step explanation:

As 2 phones are defective from a group of 98 that were checked, you can determine a percentage:

2/98= 2%

This indicates that 2% of the phones are likely to be defective and because of that you can find 2% of 850:

850*2%= 17

According to this, 17 phones are likely to be defective.

3 0
2 years ago
A study was performed on green sea turtles inhabiting a certain area.​ Time-depth recorders were deployed on 6 of the 76 capture
polet [3.4K]

Answer:

One can be 99% confident the true mean shell length lies within the above interval.

The population has a relative frequency distribution that is approximately normal.

Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.  

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

                    P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean shell length = 51.3 cm

             s = sample standard deviation = 6.6 cm

             n = sample of turtles = 6

             \mu = true mean shell length

Now, the 99% confidence interval for \mu =  \bar X \pm t_(_\frac{\alpha}{2}_)  \times \frac{s}{\sqrt{n} }

Here, \alpha = 1% so  (\frac{\alpha}{2}) = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> \mu  =  51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }

                                                         = [51.3 - 10.864 , 51.3 + 10.864]

                                                         = [40.44 cm, 62.16 cm]

The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;

C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

3 0
1 year ago
Weekly demand for a particular item averages 30 units, with a standard deviation of 4. This item is managed with a fixed-order-i
aivan3 [116]

Answer:

(B)93

Explanation:

Since we are using a fixed-order-interval model,

The Amount to Order=Expected Demand During protection Interval+Safety Stock-Amount at Hand

=d(OI+LT)+z\sigma_{d}\sqrt{OI+LT}-A

Where:  

d=weekly demand

OI=Order Interval

LT=Lead Time

z=Standard Deviation of Desired Service Level

\sigma_{d}=Standard Deviation of weekly Demand

A= Amount at Hand

=d(OI+LT)+z\sigma_{d}\sqrt{OI+LT}-A

[30(3 + 1)] + [1.964*4*\sqrt{3 + 1} - 43

= 120 + 15.712 - 43 = 92.7

4 0
1 year ago
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