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4 years ago

Which is the correct way to write three and one-tenth of a milliliter as an Arabic number?

1 answer:

4 0

we have that
three and one-tenth------> 3 1/10--------> (3*10+1)/10--------> 31/10-----> 3.1

the answer is 3.1 ml

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Trimble Graphic Design receives $2,250 from a client billed in a previous month for services provided. Which of the following ge

krok68 [10]

Answer:

Dr Cash/Bank $2,250

Cr Account Receivable $2,250

Step-by-step explanation:

At the time the transaction was carried out, that is the previous month, the accounting entries was a debit to account receivable (client account) and a credit to sales account. which is as shown below:

Dr Account Receivable $2,250

Cr Sales $2,250

Being sales on credit to a client.

The client account will have an outstanding debit balance of $2,250 until payment is made. once payment is made, the following entries will be made to ensure no outstanding balance in the client account.

Dr Cash/Bank $2,250

Cr Account Receivable $2,250.

7 0

3 years ago

a sample of 412 adults showed that 293 of them are connected the the internet from home. Construct and interpret a 90% confidenc

Anna [14]

Answer:

CI = (0.674, 0.748)

Step-by-step explanation:

The confidence interval of a proportion is:

CI = p ± SE × CV,

where p is the proportion, SE is the standard error, and CV is the critical value (either a t-score or a z-score).

We already know the proportion:

p = 293/412

p = 0.711

But we need to find the standard error and the critical value.

The standard error is:

SE = √(p (1 − p) / n)

SE = √(0.711 × (1 − 0.711) / 412)

SE = 0.0223

To find the critical value, we must first find the alpha level and the degrees of freedom.

The alpha level for a 90% confidence interval is:

α = (1 − 0.90) / 2 = 0.05

The degrees of freedom is one less than the sample size:

df = 412 − 1 = 411

Since df > 30, we can approximate this with a normal distribution.

If we look up the alpha level in a z score table or with a calculator, we find the z-score is 1.645. That's our critical value. CV = 1.645.

Now we can find the confidence interval:

CI = 0.711 ± 0.0223 * 1.645

CI = 0.711 ± 0.0367

CI = (0.674, 0.748)

So we are 90% confident that the proportion of adults connected to the internet from home is between 0.674 and 0.748.

5 0

3 years ago

when I double my number, add four, and then subtract my number and subtract 3, I get my number plus 1

Musya8 [376]

4 because doubled it equals eight, add four it equals twelve, subtract four it equals eight, subtract three and it equals five.

3 0

3 years ago

PLEASE HELP need to know cos(F), sin(F), and tan(F) WILL GIVE BRAINLIEST

SVETLANKA909090 [29]

Answer:

Cos(F) = 5/13

Sin(F) = 12/13

Tan(F) = 12/5

Step-by-step explanation:

Soh Sine: Opposite(12) over Hypotenuse(13)

Cah Cosine: Adjacent(5) over Hypotenuse(13)

Toa Tangent: Opposite(12) over Adjacent(5)

7 0

3 years ago

If the integral of the product of x squared and e raised to the negative 4 times x power, dx equals the product of negative 1 ov

Nataly_w [17]

Answer:

$A + B + E = 32$

Step-by-step explanation:

Given

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Required

Find $A +B + E$

We have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Using integration by parts

$\int {u} \, dv = uv - \int vdu$

Where

$u = x^2$ and $dv = e^{-4x}dx$

Solve for du (differentiate u)

$du = 2x\ dx$

Solve for v (integrate dv)

$v = -\frac{1}{4}e^{-4x}$

So, we have:

$\int {u} \, dv = uv - \int vdu$

$\int\limits {x^2\cdot e^{-4x}} \, dx = x^2 *-\frac{1}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x} 2xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} - \int -\frac{1}{2}e^{-4x} xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

-----------------------------------------------------------------------

Solving

$\int xe^{-4x} dx$

Integration by parts

$u = x$ ---- $du = dx$

$dv = e^{-4x}dx$ ---------- $v = -\frac{1}{4}e^{-4x}$

So:

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} + \int e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}$

So, we have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} [ -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}]$

Open bracket

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} -\frac{x}{8}e^{-4x} -\frac{1}{8}e^{-4x}$

Factor out $e^{-4x}$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{x^2}{4} -\frac{x}{8} -\frac{1}{8}]e^{-4x}$

Rewrite as:

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{4}x^2 -\frac{1}{8}x -\frac{1}{8}]e^{-4x}$

Recall that:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{64}Ax^2 -\frac{1}{64} Bx -\frac{1}{64} E]Ce^{-4x}$

By comparison:

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

$-\frac{1}{8} = -\frac{1}{64}E$

Solve A, B and C

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

Divide by $-x^2$

$\frac{1}{4} = \frac{1}{64}A$

Multiply by 64

$64 * \frac{1}{4} = A$

$A =16$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

Divide by $-x$

$\frac{1}{8} = \frac{1}{64}B$

Multiply by 64

$64 * \frac{1}{8} = \frac{1}{64}B*64$

$B = 8$

$-\frac{1}{8} = -\frac{1}{64}E$

Multiply by -64

$-64 * -\frac{1}{8} = -\frac{1}{64}E * -64$

$E = 8$

So:

$A + B + E = 16 +8+8$

$A + B + E = 32$

4 0

3 years ago