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sashaice [31]
3 years ago
14

7.3 × 9.6= Multiply decimals

Mathematics
1 answer:
cluponka [151]3 years ago
4 0
The answer to your question is 70.08
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Elena wants to make a scaled drawing of her bedroom. Her bedroom is a rectangle with length 500 cm.She decides on a scale of 1 t
lapo4ka [179]

Answer:

Length = 200cm by 150cm

Step by step explanation:

Bringing it to actual measurements from the scale of 1 to 50, 3cm becomes 3 x 50 = 150 and 4cm becomes 4 x 50 = 200

7 0
3 years ago
Lin's father is paying for a $20 meal. He has a 15%-off coupon for the meal. After the discount, a 7% sales tax is applied.
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I’m pretty sure it’s $16.60 but I’m not sure but I hope you get it right!
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3 years ago
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Alan buys 5.3 ounces of sour patch kids candy. After sharing with his friends he returns to buy an additional 6.75 ounces. How m
Natasha_Volkova [10]

Answer:

Alan buys 12.05 ounces of sour patch kids candy in all.

Step-by-step explanation:

We are given that Alan buys 5.3 ounces of sour patch kids candy. After sharing with his friends he returns to buy an additional 6.75 ounces.

And we have to find the total ounces he buys in all.

As we know that for finding the total quantity or amount, we will use addition for calculating it.

Firstly, Alan buys = 5.3 ounces of sour patch kids candy

Additional ounces of sour patch kids candy Alan buys = 6.75

So, the total ounces of sour patch kids candy he buys in all = 5.3 + 6.75

                                                                                                   = 12.05 ounces

Hence, he buys 12.05 ounces in all.

7 0
3 years ago
A random sample of the actual weight of 5-lb bags of mulch produces a mean of 4.963 lb and a standard deviation of 0.067 lb. If
dsp73

Answer: B) 4.963±0.019.

Step-by-step explanation:

Confidence interval for population mean ( when population standard deviation is not given) is given by :-

\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}

, where \overline{x} = Sample  mean

n= Sample size

s= sample standard deviation

t* = critical t-value.

As per given:

n= 50

Degree of freedom = n-1 =49

\overline{x}= 4.963\ lb

s= 0.067 lb

For df = 49 and significance level of 0.05 , the critical two-tailed t-value ( from t-distribution table) is 2.010.

Now , substitute all values in the formula , we get

4.963\pm (2.010)\dfrac{0.067}{\sqrt{50}}\\\\ 4.963\pm (2.010)(0.0094752)\\\\ 4.963\pm0.019045152\approx4.963\pm0.019

Hence,  a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company is 4.963\pm0.019.

Thus , the correct answer is B) 4.963±0.019.

7 0
3 years ago
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