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Sauron [17]
2 years ago
11

A local newspaper did a survey asking people of different ages how many hours they spend using social media sites each week. The

results of the survey and the least squares regression line are displayed in the graph. Use the least squares regression line to predict about how many hours a 40-year-old person spends on social media sites each week.
Mathematics
1 answer:
FrozenT [24]2 years ago
5 0

Answer:

Step-by-step explanation:

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The sum of two numbers is 64 . the larger number is 18 more than the smaller number. what are the numbers?
-BARSIC- [3]
64-18 is 46
46/2 is 23
23+18 is 41

Your #s are 21 and 43
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Find the sum.<br> 2y<br> y2 + 2y + 1<br> +4/<br> y2 + 3y + 2
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Answer:

did u put the equation right?

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3 years ago
Jake is going to the store to buy candles. Small candles cost $3.50 and
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Answer:

Let

S= #small candles

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S+ L =>20

3.50S + 5.00L < 80

Step-by-step explanation:

3 0
3 years ago
How to solve -2p^2=16p+24
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Add  2p²  to each side of the equation.  Then you have

                 2p² + 16p + 24  =  0 .

Before you roll up your sleeves and start working on it, you can make it
even more convenient if you divide each side by  2 .  Then you have:

                 p² + 8p + 12 = 0 .

Now you have a nice, comfortable, familiar-looking quadratic equation. 
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8 0
3 years ago
Two 5-year girls, Alyse and Jocelyn, have been training to run a 1-mile race. Alyse's 1 mile time A is approximately Normally di
tatyana61 [14]

Answer:

1.7 × 10⁻⁴

Step-by-step explanation:

The question relates to a two sample z-test for the comparison between the means of the two samples  

The null hypothesis is H₀:  μ₁ ≤ μ₂

The alternative hypothesis is Hₐ: μ₁ > μ₂

z=\dfrac{(\bar{x}_1-\bar{x}_2)-(\mu_{1}-\mu _{2} )}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}

Where;

\bar {x}_1 = 13.5

\bar {x}_2 = 12

σ₁ = 2.5

σ₂ = 1.5

We set our α level at 0.05

Therefore, our critical z = ± 1.96

For n₁ = n₂ = 23, we have;

z=\dfrac{(13.5-12)-(0)}{\sqrt{\dfrac{2.5^{2} }{23}-\dfrac{1.5^{2}}{23}}} = 3.5969

We reject the null hypothesis at α = 0.05, as our z-value, 3.5969 is larger than the critical z, 1.96 or mathematically, since 3.5969 > 1.96

Therefore, there is enough statistical evidence to suggest that Alyse time is larger than Jocelyn in a 1 mile race on a randomly select day and the probability that Alyse has a larger time than Jocelyn is 0.99983

Therefore;

The probability that Alyse has a smaller time than Jocelyn is 1 - 0.99983 = 0.00017 = 1.7 × 10⁻⁴.

8 0
3 years ago
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