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Maslowich
3 years ago
13

Can someone help me with 10 and 11 pls

Mathematics
1 answer:
Dmitriy789 [7]3 years ago
8 0

Answer:

Answer for 10: 7 Solutions Answer for 11: 7-28= y7= x7

Step-by-step explanation:

Read above

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Between which two consecutive integers does <img src="https://tex.z-dn.net/?f=%5Csqrt%7B%7D%2012%5C%5C" id="TexFormula1" title="
Annette [7]

The most appropriate choice for square root will be given by-

\sqrt{12} lies between 3 and 4

What is square root of a number?

Number raised to the power \frac{1}{2} is the square root of  the number.

Here

\sqrt{12}\\ 2\sqrt{3}\\ 2*1.732\\3.464\\

3.464 lies between 3 and 4

\sqrt{12} lies between 3 and 4

To learn more about square root of a number, refer to the link:

brainly.com/question/3617398

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7 0
11 months ago
Ding ding ding help is now!
Ivenika [448]

Answer:

I believe the missing number is 65

Step-by-step explanation:

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2 years ago
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What is 0.32 written as a decimal
vredina [299]
Decimal: .32
fraction: 32/100 which, when reduced, is 8/25 
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3 years ago
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Does the following relation on x and y make for a function of x?
Mkey [24]
Yes, because by the definition of function, for every single input there is one out put corresponding to that.
3 0
3 years ago
Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, 20, percent of the
Arturiano [62]

Answer:

Step-by-step explanation:

From the given information:

We can compute the  null hypothesis & the alternative hypothesis as:

{H_o}:\text{Distribution of snowfalls in her hometown is similar to claimed percentage }

{H_a}:\text{Distribution of snowfalls in her hometown is not similar to claimed percentage }

The degree of freedom = n - 1

The degree of freedom = 5 - 1

The degree of freedom = 4

At the level of significance of 0.05 and degree of freedom 4,

the rejection region = 9.488

However, we can compute the chi-square X² goodness of fit test as:

   

months  frequency (p)  observed O Expected E  Chi-square X^2= \dfrac{(O-E)^2}{E}

Dec          0.2                  16                   16                \dfrac{(16-16)^2}{16} =0      

Jan           0.250             11                   20                \dfrac{(11-20)^2}{20} =4.050      

Feb           0.200             16                  16                 \dfrac{(16-16)^2}{16} =0      

Mar           0.200             18                  16                 \dfrac{(18-16)^2}{16} =0.250

Apr           0.150               19                  12                 \dfrac{(19-12)^2}{12} =4.083

Total            1.000           80                 80                                  8.3833    

∴

The test statistics X² = 8.3833

Thus; we fail to reject the H_o since test statistics X² doesn't fall in the rejection region.

Therefore; there is sufficient evidence to conclude that the distribution of snowfalls in her hometown is not similar to the claimed percentage.

6 0
3 years ago
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