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harina [27]
3 years ago
9

Convert 120 mins in to hours

Mathematics
2 answers:
Firlakuza [10]3 years ago
7 0

Answer:

2 hours

Step-by-step explanation:

1 hour = 60

60 x 2 = 120

Zanzabum3 years ago
6 0
The answer would be two hours, one hour is 60 minutes so 120/60 would be 2
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The area of a square garden is 49 ft^2. How long is each side of the​ garden?
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7

Step-by-step explanation:

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4 years ago
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For f(x) = 3x^2 – X+5, 2f (-3) - f(4)=
Alenkinab [10]

Answer:

3x^2-x+5.2f\left(-3\right)-f\left(4\right) =

3x^2-x-19.6f

Step-by-step explanation:

3x^2-x+5.2f\left(-3\right)-f\left(4\right)

=3x^2-x-5.2f\cdot \:3-f\cdot \:4

=3x^2-x-15.6f-4f

=3x^2-x-19.6f

8 0
3 years ago
Let a,b,c ∈ Z. Define the highest common factor hcf(a,b,c) to be the largest positive integer that divides a,b and c. Prove that
kakasveta [241]

21 they have too be right next to each other and threy have too be right

8 0
3 years ago
A circle is enclosed in a square. Find the area of the shaded region. Use pi = 3.14 10m O 25 sq. cm. 100 sq. cm O 78.5 sq. cm O
Juli2301 [7.4K]

Answer:

25 sq. cm

Step-by-step explanation:

i think so i'm not sure if it is correct

6 0
3 years ago
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A university researcher wants to estimate the mean number of novels that seniors read during their time in college. An exit surv
lys-0071 [83]

Answer:

Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

Step-by-step explanation:

Given - A university researcher wants to estimate the mean number

            of  novels that seniors read during their time in college. An exit

            survey was conducted with a random sample of 9 seniors. The

            sample mean was 7 novels with standard deviation 2.29 novels.

To find - Assuming that all conditions for conducting inference have

              been met, which of the following is a 94.645% confidence

              interval for the population mean number of novels read by

              all seniors?

Proof -

Given that,

Mean ,x⁻ = 7

Standard deviation, s = 2.29

Size, n = 9

Now,

Degrees of freedom = df

                                = n - 1

                                = 9 - 1

                                = 8

⇒Degrees of freedom = 8

Now,

At 94.645% confidence level

α = 1 - 94.645%

   =1 - 0.94645

  =0.05355 ≈ 0.05

⇒α = 0.5

Now,

\frac{\alpha}{2} = \frac{0.05}{2}

  = 0.025

Then,

t_{\frac{\alpha}{2}, df }  = 2.306

∴ we get

Population mean = x⁻ ± t_{\frac{\alpha}{2}, df } ×\frac{s}{\sqrt{n} }

                           = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

⇒Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

3 0
3 years ago
Read 2 more answers
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