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3 years ago

Last month we had only five sunny days. There were 30 days in all last month.

Mathematics

2 answers:

allochka39001 [22]3 years ago

7 0

Answer:

Step-by-step explanation:

it is 6

neonofarm [45]3 years ago

5 0

Answer:

Step-by-step explanation:

30 ÷ 5 = 6

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Item 9 Evaluate (3−1) 3 +7(6)− 5 2 (3−1)3+7(6)−52 .

krok68 [10]

The answer this problem is -247

6 0

3 years ago

Help: I am a number between 60 and 100. My ones digit is six less than my tens digit. I am a prime number.

Reptile [31]

Answer:

Let's name the digit:

x- ones digit

y - tens digit

we know that x=y-2.

Now, y can be 6,7,8,9 (the number is between 60 and 100 (so depending on your understanding of "between", 0 is also possible. but then the number would have to have -2 as its ones digit, so in any case, it's not possible).

So the possibilities with x=y-2:

Out of those 64 and 86 are even, so they can't be prime.

75 has 5 in its ones number: it's divisible by 5.

so the correct answer is 97.

Step-by-step explanation:

Hope this helps!

5 0

3 years ago

Read 2 more answers

What are the chances your flight will leave on time? Below are a histogram and summary statistics for the percentage of flights

Jet001 [13]

Answer:

80.0456< $\bar x$ <81.1210

Step-by-step explanation:

-Given the mean, $\bar y=80.5833$ and $s=2.77369$ , the confidence interval can be calculated using the formula:

$\bar x\pm z\times \frac{\sigma}{\sqrt{n}}$

#We substitute our values in the formula to solve for CI:

$=\bar x\pm z\times \frac{\sigma}{\sqrt{n}}\\\\=\bar y\pm z_{0.05}\times \frac{s}{\sqrt{72}}\\\\=80.5833\pm 1.645\times \frac{2.77369}{\sqrt{72}}\\\\=80.5833\pm0.5377\\\\=[80.0456,81.1210]$

Hence, the confidence interval lies between 80.0456 and 81.1210

4 0

3 years ago

Find the marginal and average revenue functions associated with the demand function P= -0.3Q + 221

LekaFEV [45]

Answer:

Marginal revenue = R'(Q) = -0.6 Q + 221

Average revenue = -0.3 Q + 221

Step-by-step explanation:

As per the question,

Functions associated with the demand function P= -0.3 Q + 221, where Q is the demand.

Now,

As we know that the,

Marginal revenue is the derivative of the revenue function, R(x), which is equals the number of items sold,

Therefore,

R(Q) = Q × ( -0.3Q + 221) = -0.3 Q² + 221 Q

∴ Marginal revenue = R'(Q) = -0.6 Q + 221

Now,

Average revenue (AR) is defined as the ratio of the total revenue by the number of units sold that is revenue per unit of output sold.

$Average\ revenue\ = \frac{Total\ revenue}{number\ of\ units\ sold}$

Where Total Revenue (TR) equals quantity of output multiplied by price per unit.

TR = Price (P) × Total output (Q) = (-0.3Q + 221) × Q = -0.3 Q² + 221 Q

$Average\ revenue\ = \frac{TR}{Q}$

$Average\ revenue\ = \frac{-0.3Q^{2}+221Q}{Q}$

∴ Average revenue = -0.3Q + 221

5 0

3 years ago

The sum of a 3 digit number and a 1 digit number is 217. The product of the numbers is 642. If one number is between 200 and 225

algol [13]

Let, the numbers be "x" and "y". Consider "x" as the triple digit number.
Now, according to the question,
x + y = 217..........................equation (1)

x * y = 642............................equation (2)

Now,
Taking equation (2),
x * y = 642
y = 642 / x..................................equation (3)

Now, Taking equation (1),
$x+y=217$

Substituting the value of y from equation (3), we get,

$x+ \frac{642}{x} =217$

$\frac{x *x+642}{x} =217$

$x^{2} +642 =217*x$

$x^{2} +642 =217x$

$x^{2} +642-217x =0$

$x^{2} -217x+642 =0$

$x^{2} -3x-214x+642 =0$

$x(x-3)-214(x-3)=0$

$(x-3)(x-214)=0$

Using zero product property,
EITHER,
x - 3 = 0
x = 3
OR,
x - 214 = 0
x = 214
Since, "x" is the triple digit number, x = 214.
Now,
Taking equation (2),
x * y = 642
Substituting the value of "x" in the equation, we get,
(214) * y = 642
y = 642 / 214
y = 3

So, the numbers are 214 and 3.

5 0

3 years ago