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AnnyKZ [126]
2 years ago
15

You are signing up for a new gym on New Year's Day. Two gyms memberships have different payment plans. Gym A has a $60 signup fe

e and a monthly fee of $10. Gym B has a signup fee of $20 and a monthly fee of $15. At which month would the membership costs be the same?
Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
6 0

Answer: August or the 8th month

Step-by-step explanation:

Assuming the month that the costs will be the same is x.

Gym A cost function will be = 60 + 10x

Gym B cost function will be =  20 + 15x

The formula will therefore be;

60 + 10x = 20 + 15x

60 - 20 = 15x - 10x

40 = 5x

x = 8

8th month is August.

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At a large company, employees can take a course to become certified to perform certain tasks. There is an exam at the end of the
WINSTONCH [101]

Answer:

A Type II error is when the null hypothesis is failed to be rejected even when the alternative hypothesis is true.

In this case, it would represent that the new program really increases the pass rate, but the sample taken is not enough statistical evidence to prove it. Then, the null hypothesis is not rejected.

The consequence is that the new method would be discarded (or changed) eventhough it is a real improvement.

Step-by-step explanation:

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2 years ago
Two cones are similar. The surface area of the larger cone is 65π square inches. The surface area of the smaller cone is 41.6π s
vaieri [72.5K]
8 inches hope it helps

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2 years ago
Read 2 more answers
A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y de
strojnjashka [21]

Answer:

a) P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

a= μ-3.16*σ , b= μ+3.16*σ

b) P(Y≥ μ+3*σ ) ≥ 0.90

b= μ+3*σ

Step-by-step explanation:

from Chebyshev's inequality for Y

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

where

Y =  the number of fish that need be caught to obtain at least one of each type

μ = expected value of Y

σ = standard deviation of Y

P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean

k= parameter

thus for

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

P{a≤Y≤b} ≥ 0.90 →  1-1/k² = 0.90 → k = 3.16

then

P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

using one-sided Chebyshev inequality (Cantelli's inequality)

P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)

P{Y≥b} ≥ 0.90  →  1- σ²/(σ²+λ²)=  1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ

then for

P(Y≥ μ+3*σ ) ≥ 0.90

5 0
2 years ago
F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa
DerKrebs [107]

a. Parameterize C by

\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath

with 0\le t\le1.

b/c. The line integral of \vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath over C is

\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt

=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt

=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt

=\displaystyle10\int_0^1\mathrm dt=\boxed{10}

d. Notice that we can write the line integral as

\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)

By Green's theorem, the line integral is equivalent to

\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy

where D is the triangle bounded by C, and this integral is simply twice the area of D. D is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0
2 years ago
The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students re
Vladimir79 [104]

Answer:

The 95% of confidence intervals

(2.84 ,2.99)

Step-by-step explanation:

A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16

given small sample size n =20

sample mean x⁻ =2.92

sample standard deviation 'S' =0.16

level of significance ∝ =  0.95

The 95% of confidence intervals

x^{-}  ± t_{\alpha } \frac{S}{\sqrt{n} }

the degrees of freedom γ=n-1 =20-1=19

t-table 2.093

(x^{-}  - t_{\alpha } \frac{S}{\sqrt{n} },x^{-}  + t_{\alpha } \frac{S}{\sqrt{n} })

(2.92 - 2.093(\frac{0.16}{\sqrt{20} } ,2.92+2.093(\frac{0.16}{\sqrt{20} } )

(2.92-0.0748,2.92+0.0748)

(2.84 ,2.99)

Therefore the 95% of confidence intervals

(2.84 ,2.99)

4 0
2 years ago
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