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DedPeter [7]
3 years ago
6

Total cost 26 oz at $0.15 per oz

Mathematics
2 answers:
RoseWind [281]3 years ago
8 0
$0.15 per ounce * 26 Oz = $3.90
alexandr402 [8]3 years ago
6 0
26*0.15=3.90

if you want 26 ounces of something at $0.15 per ounce, then you will pay $3.90

(plus tax)
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Which statement are true based on the diagram? Select 2 options
Step2247 [10]

Answer:

second and last are true

Step-by-step explanation:

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4 years ago
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A film ends at 1.05 a.m and lasts for 1 hour 45 min. At what time does the film start?
Ad libitum [116K]

u will have to add the two

1hr 45mins + 1:05

remember add hrs to hrs and minutes to minutes

1+1 =2

45+5=50

which means the movie started at 2:50 am

and to verify ur answer subtract 2:50 from 1:05

= 1hrs 45 mins

<em>Answer</em><em>=</em><em> </em><em>2</em><em>:</em><em>5</em><em>0</em><em> </em><em>am</em><em> </em>

6 0
3 years ago
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2. Ravi purchased an old house for Rs. 76.5000 and
lorasvet [3.4K]

Answer:

Rs. 924000.

Step-by-step explanation:

Cost of  house = 765000

Additional money spent on it = 115000

Total cost incurred by Ravi = 765000 + 115000 = 880000

Gain  = 5% of total cost

gain in Rs = 5/100 * 880000 = Rs. 44000

Total selling price of house = total cost incurred + profit = 880000+ 44000

Total selling price of house = Rs. 924000

Thus, Ravi got  Rs. 924000.

3 0
4 years ago
A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken
mihalych1998 [28]

Answer:

n=6147

Step-by-step explanation:

1) Notation and definitions

X=1 number of citrus trees that are infected with the citrus red mite.

n=5 random sample taken

\hat p=\frac{1}{5}=0.2 estimated proportion of citrus trees that are infected with the citrus red mite.

p true population proportion of citrus trees that are infected with the citrus red mite.

Me=0.01 represent the margin of error desired

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical values would be given by:

t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.2(1-0.2)}{(\frac{0.01}{1.96})^2}=6146.56  

And rounded up we have that n=6147

5 0
4 years ago
Need help ASAP please
inysia [295]

Answer:

m=1

Step-by-step explanation:

y2-y1/x2-x1

3 0
3 years ago
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