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4 years ago

425 is decreased to 319

1 answer:

5 0

425 is decreased to 319.

"Decreased to" and "decreased by" have different meaning.
"decreased to" means that the number given after the phrase is the difference in the subtraction.
"decreased by" means that the number given after the phrase is the number deducted from the other number to get the difference.

319 is the difference. We need to get the number to be deducted from 425.

425 - x = 319
425 - 319 = x
106 = x

to check:
425 - 106 = 319
319 = 319

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Sarah Earns a weekly salary of 445+ commission of 8.5% on sale at the clothing store when she works how much should she turned i

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3 years ago

A random sample of 20 married women showed that the mean time spent on housework by them was 29.8 hours a week with a standard d

DanielleElmas [232]

Answer:

95% confidence interval for the mean time spent on housework per week by all married women.

( 26.66 , 32.94)

Step-by-step explanation:

Step(i):-

Given random sample size 'n' = 20

Mean of the sample (x⁻ ) = 29.8 hours

Standard deviation of the sample (S) = 6.7

Given Margin of error = 3.14

Step(ii):-

95% confidence interval for the mean is determined by

$(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } , x^{-} +t_{0.05} \frac{S}{\sqrt{n} })$

We know that margin of error is determined by

$M.E = \frac{t_{0.05}XS.D }{\sqrt{n} } = 3.14$

Now 95% confidence interval for the mean time spent on housework per week by all married women.

$(29.8 - 3.14 , 29.8+3.14)$

( 26.66 , 32.94)

8 0

3 years ago

Write a real-world problem that can be represented by the equation f + 5= -9.

miv72 [106K]

Answer: Katie is scuba diving at an altitude of -14 ft below sea level. She goes up 5 ft. What altitude is Katie at after she goes up 5 ft?

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Calculus 2. Please help

Anarel [89]

Answer:

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}}} \, dx = \infty$

General Formulas and Concepts:

Algebra I

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integrals

Definite Integrals

Integration Constant C

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

U-Solve

Improper Integrals

Exponential Integral Function: $\displaystyle \int {\frac{e^x}{x}} \, dx = Ei(x) + C$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \int\limits^1_0 {\frac{e^{-x^2}}{x} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \int\limits^1_a {\frac{e^{-x^2}}{x} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set: $\displaystyle u = -x^2$
Differentiate [Basic Power Rule]: $\displaystyle \frac{du}{dx} = -2x$
[Derivative] Rewrite: $\displaystyle du = -2x \ dx$

Rewrite u-substitution to format u-solve.

Rewrite du: $\displaystyle dx = \frac{-1}{2x} \ dx$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {-\frac{e^{-x^2}}{x} \, dx$
[Integral] Substitute in variables: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {\frac{e^{u}}{-2u} \, du$
[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}\int\limits^1_a {\frac{e^{u}}{u} \, du$
[Integral] Substitute [Exponential Integral Function]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(u)] \bigg| \limits^1_a$
Back-Substitute: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-x^2)] \bigg| \limits^1_a$
Evaluate [Integration Rule - FTC 1]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-1) - Ei(a)]$
Simplify: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{Ei(-1) - Ei(a)}{2}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \infty$

∴ $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$ diverges.

Topic: Multivariable Calculus

7 0

3 years ago

A large industrial firm allows a discount on any invoice that is paid within 30 days. of all invoices, 10% receive the discount.

Daniel [21]

This situation has two outcomes: either an invoice is discounted or not. This two outcomes satisfy what we call a binomial distribution (note "bi" in binomial).

The binomial distribution tells us the probability that a randomly selected sample will have the outcome of success. In this case, we consider receiving the discount as the "success" outcome while not receiving it means "failure". The distribution takes the form:

$P(X=x)= _{n}C_{x} p^{x} q^{n-x}$
where n is the total number of samples, x is the number of samples you'll expect to have a success outcome, p is the probability of success, q is the probability of failure, and nCx is the combination of n samples taken x at a time.

You have an inconsistency regarding the total number of samples by the way, but I will take the first mentioned sample of 15 as I answer (you can just follow through the same process for another value).

For the next step, let's digest the problem to get the needed variables.

The total number of samples, n, is equal to 15. The probability of success (receiving a discount), p, is 10% or 0.1, and the probability of failure (not receiving a discount) is 90% or 0.9.

For P(X=x), we need to find the probability that less than two of the samples have success outcomes therefore we need to find the probability that NO invoice will receive the discount plus the probability that ONE invoice will receive the discount. This is equivalent to saying
$P(X\ \textless \ 2)=P(X=0)+P(X=1)$

Calculating these probabilities we'll get:
$P(X=0)= _{15}C_{0} (0.1)^{0} (0.9)^{15}=0.206$
$P(X=1)= _{15}C_{1} (0.1)^{1} (0.9)^{14}=0.343$
$P(X\ \textless \ 2)=P(X=0)+P(X=1)=0.206+0.343=0.549$

ANSWER: The probability that fewer than 2 sampled invoices will receive the discount is 0.549 or 54.9%.

6 0

3 years ago