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3 years ago

0.9/0.5 = a/10 Plz help

Mathematics

2 answers:

MissTica3 years ago

8 0

Answer:

a = 18

Step-by-step explanation:

Step 1: Cross-multiply.

$0.5 * a = 10*0.9$
$0.5a =9$

Step 2: Divide both sides by 0.5

$\frac{0.5a}{0.5} =\frac{9}{0.5}$
$a = \frac{90}{5}$
$a = 18$

Step 3: Check if solution is correct.

$\frac{0.9}{0.5}=\frac{18}{10}$
$\frac{9}{5} =\frac{18}{10}$
$\frac{18/2}{10/2} =\frac{9}{5}$
$\frac{9}{5} =\frac{9}{5}$

Therefore, a = 18.

jonny [76]3 years ago

7 0

Answer:

Step-by-step explanation:

10/0.5 = 20

20 * 0.9 = 18

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The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviati

Stolb23 [73]

Answer:

(a) less than 10 minutes

= 0.5

(b) between 5 and 10 minutes

= 0.5

Step-by-step explanation:

We solve the above question using z score formula. We given a random number of samples, z score formula :

z-score is z = (x-μ)/ Standard error where

x is the raw score

μ is the population mean

Standard error : σ/√n

σ is the population standard deviation

n = number of samples

(a) less than 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5

(b) between 5 and 10 minutes

i) For 5 minutes

x = 5 μ = 10, σ = 2 n = 50

z = 5 - 10/2/√50

z = -5 / 0.2828427125

= -17.67767

P-value from Z-Table:

P(x<5) = 0

Using the z table to find the probability

P(z ≤ 0) = P(z = -17.67767) = P(x = 5)

= 0

ii) For 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is

P(x = 10) - P(x = 5)

= 0.5 - 0

= 0.5

3 0

3 years ago

Find the payment necessary to amortize a 6.3% loan of 7500 compounded semiannually, with 6 semiannual payments. Find the payment

Lorico [155]

Answer:

Semiannual payment = $ 1391.37

Total payment = $ 8348.22

Interest paid = $ 848.22

Step-by-step explanation:

Since, the semiannual payment formula of a loan,

$P=\frac{PV(\frac{r}{2})}{1-(1+\frac{r}{2})^{-n}}$

Where,

PV = present value of the loan,

n = number of semiannual payments,

r = annual rate,

Here, PV = 7500, r = 6.3% =0.063, n = 6,

By substituting the value,

The semiannual payment would be,

$P=\frac{7500(\frac{0.063}{2})}{1-(1+\frac{0.063}{2})^{-6}}$

$\approx \$ 1391.37$

Also, total payment = semiannual payment × total semiannual periods

= 1391.37 × 6

= $ 8348.22,

Also, the interest paid = total payment - present value

= 8348.22 - 7500

= $ 848.22

6 0

3 years ago

1. Write as a base with a positive exponent, then evaluate. Write your final answer as a

vodomira [7]

Answer:

Well a base with a positive exponent would be 5^3. All this means is 5x5x5 which is 125.

Step-by-step explanation:

Hope this helps!

5 0

3 years ago

What is the equation of an asymptote of the hyperbola whose equation is (x-2)^2/4 - (y-1)^2/36 = 1?

UNO [17]

Answer: it’s C. Y=3x-5

4 0

3 years ago

A fund earns a nominal rate of interest of 6\% compounded every two years. Calculate the amount that must be contributed now to

Nesterboy [21]

Answer:

$712.

Step-by-step explanation:

We have been given that a fund earns a nominal rate of interest of 6% compounded every two years. We are asked to find the amount that must be contributed now to have 1000 at the end of six years.

We will use compound interest formula to solve our given problem.

$A=P(1+\frac{r}{n})^{nt}$ , where,

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

$6\%=\frac{6}{100}=0.06$

Since interest is compounded each two years, so number of compounding per year would be 1/2 or 0.5.

$1000=P(1+\frac{0.06}{0.5})^{0.5*6}$

$1000=P(1+0.12)^{3}$

$1000=P(1.12)^{3}$

$1000=P*1.404928$

$\frac{1000}{1.404928}=\frac{P*1.404928}{1.404928}$

$P=711.7802478$

$P\approx 712$

Therefore, an amount of $712 must be contributed now to have 1000 at the end of six years.

7 0

3 years ago