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aniked [119]
3 years ago
6

The distance from the school to the library is 12 miles. The distance from the school to the park is 75% of this distance.

Mathematics
2 answers:
strojnjashka [21]3 years ago
8 0
75% of 12 can be represented with the equation 75/100=x/12

Then to solve
75(12)= 100x
900 = 100x
9=x

So the distance from the school to the park is 9 miles.
Varvara68 [4.7K]3 years ago
7 0

Answer:

9

Step-by-step explanation:

all I just did was add the 7 and 2 but turned the 70 into a 7

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2 years ago
8.6 mi<br> 18 mi<br> A) 53.2 mi?<br> C) 106.4 mi?<br> E) 212.8 mi?<br> B) 221 mi?<br> D) 110.5 mi?
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4 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

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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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