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3 years ago

3 1/2 x 1 1/3 Please explain how to get the answer

Mathematics

2 answers:

Nataly_w [17]3 years ago

8 0

Answer:

4 2/3

Step-by-step explanation:

You need to turn the fractions into improper fractions

To do that you multiply the whole number by the denominator and add the numerator, doing that will give you your new numerator, but the denominator stays the same

3 1/2. 3×2 = 6. 6+1=7 3 1/2 = 7/2

1 1/3. 1×3 = 3. 3+1 = 4. 1 1/3 = 4/3

Then multiply both numerator by each other and do the same for the denominators

7×4 = 28

2×3 = 6

28/6

Then simply it

28/6 = 4 2/3

Colt1911 [192]3 years ago

7 0

Answer: 77/3 or 25.67

Step-by-step explanation:

3 1/2 * 11/3

3 * 2+1=7

7/3 * 11/3

7 * 11= 77

3 stays the same

77/3 or 25.67

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(9x + 7)(4x + 1)(3x + 4) = 0 1 root 3 roots 4 roots 9 roots

asambeis [7]

Answer:

3 roots

Step-by-step explanation:

(9x+7)(4x+1)(3x+4)=0

x=-7/9 or - 1/4 or - 4/3

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3 years ago

in a pottery class, a teacher has 2/3 pound of clay for 6 students. If she gives each student a equal amount of clay, how much w

Jlenok [28]

She will give 1/9 pound of clay to each student

8 0

3 years ago

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If a is the average (arithmetic mean) of 4x and 7, b is the average of 5x and 6, and c is the average of 3x and 11, what is the

Debora [2.8K]

Answer:

2x+4

Step-by-step explanation:

Given that a is the average (arithmetic mean) of 4x and 7, b is the average of 5x and 6, and c is the average of 3x and 11

By definition of arithmetic mean,

we have $a=\frac{4x+7}{2} \\b=\frac{5x+6}{2} \\c=\frac{3x+11}{2}$

Let us calculate sum of a,b,c in terms of x

a+b+c = $\frac{4x+7+5x+6+3x+11}{2} \\=6x+12$

Average = sum/3 = $\frac{6x+12}{3} =2x+4$

7 0

4 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago

How many inches is 2 1/6 ft

forsale [732]

Answer:

26 inches

Step-by-step explanation:

8 0

3 years ago

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