Ask question

Ask question

All categories

sertanlavr [38]

3 years ago

11

death valley is the hottest place in the united states. the highest temperature ever recorded there was 134F the lowest temperat

ure recorded there was 15F write two inequalities that would describe the temperature, in degrees fahrenheit, in death valley at any time since temperatures have been recorded

2 answers:

vampirchik [111]3 years ago

6 0

Answer:

x ≤ 134 F

x ≥ 15 F

I am Lyosha [343]3 years ago

4 0

The hottest air temperature ever recorded in Death Valley (Furnace Creek) was 134°F (57°C) on July 10, 1913. During the heat wave that peaked with that record, five consecutive days reached 129° F (54°C) or above. Death Valley holds the record for the hottest place on earth.

You might be interested in

The measure of an angle is 142.5°. What is the measure of its supplementary angle?

Debora [2.8K]

Answer:

37.50°

Step-by-step explanation:

The sum of both sides of an angle is 180°.

180° - 142.5° = 37.50°

7 0

3 years ago

there are 495 students in a senior class. The cost of a senior trip is $120 per student. write an expression that can represent

olga55 [171]

$120 per each student x 495 students = $59,400
Hope this helps!

6 0

4 years ago

Read 2 more answers

When writing a number in scientific notation, if you move the decimal point to the left 3 places, then you must _____. multiply

Naddika [18.5K]

Answer:

Multiply by 10.

Step-by-step explanation:

When writing a number in scientific notation, if you move the decimal point to the left 3 places, then you must multiply by 10.

6 0

3 years ago

If you eat a diet with 2,000 kilocalories and 45 percent of those calories come from protein, about how many grams of protein di

jekas [21]

The amount of Protein intake is 900 kilocalorie.

We have - 2000 Kilocalories of diet and 45 Percent of calories come from Proteins.

Wе have to find out the amount of protein intake.

<h3>If ' x ' students out of total ' n ' students in a class are infected by virus. Then, what percentage of students are infected.</h3>

Percentage of students affected are - $\frac{x}{n} \times 100$

In the question given -

Let the amount of Protein intake be y.

Then -

45 = y % of 2000

45 = $\frac{y}{2000} \times100$

y = $\frac{45\times2000}{100}$

y = 900 Kilocalorie

Hence, the amount of Protein intake is 900 kilocalorie.

To solve more questions on Percentage, visit the link below-

brainly.com/question/14801224

#SPJ1

7 0

2 years ago

Canadians who visit the United States often buy liquor and cigarettes, which are much cheaper in the United States. However, the

fenix001 [56]

Answer:

(a): Marginal pmf of x

$P(0) = 0.72$

$P(1) = 0.28$

(b): Marginal pmf of y

$P(0) = 0.81$

$P(1) = 0.19$

(c): Mean and Variance of x

$E(x) = 0.28$

$Var(x) = 0.2016$

(d): Mean and Variance of y

$E(y) = 0.19$

$Var(y) = 0.1539$

(e): The covariance and the coefficient of correlation

$Cov(x,y) = 0.0468$

$r \approx 0.2657$

Step-by-step explanation:

Given

<em>x = bottles</em>

<em>y = carton</em>

<em>See attachment for complete question</em>

<em />

Solving (a): Marginal pmf of x

This is calculated as:

$P(x) = \sum\limits^{}_y\ P(x,y)$

So:

$P(0) = P(0,0) + P(0,1)$

$P(0) = 0.63 + 0.09$

$P(0) = 0.72$

$P(1) = P(1,0) + P(1,1)$

$P(1) = 0.18 + 0.10$

$P(1) = 0.28$

Solving (b): Marginal pmf of y

This is calculated as:

$P(y) = \sum\limits^{}_x\ P(x,y)$

So:

$P(0) = P(0,0) + P(1,0)$

$P(0) = 0.63 + 0.18$

$P(0) = 0.81$

$P(1) = P(0,1) + P(1,1)$

$P(1) = 0.09 + 0.10$

$P(1) = 0.19$

Solving (c): Mean and Variance of x

Mean is calculated as:

$E(x) = \sum( x * P(x))$

So, we have:

$E(x) = 0 * P(0) + 1 * P(1)$

$E(x) = 0 * 0.72 + 1 * 0.28$

$E(x) = 0 + 0.28$

$E(x) = 0.28$

Variance is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Calculate $E(x^2)$

$E(x^2) = \sum( x^2 * P(x))$

$E(x^2) = 0^2 * 0.72 + 1^2 * 0.28$

$E(x^2) = 0 + 0.28$

$E(x^2) = 0.28$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 0.28 - 0.28^2$

$Var(x) = 0.28 - 0.0784$

$Var(x) = 0.2016$

Solving (d): Mean and Variance of y

Mean is calculated as:

$E(y) = \sum(y * P(y))$

So, we have:

$E(y) = 0 * P(0) + 1 * P(1)$

$E(y) = 0 * 0.81 + 1 * 0.19$

$E(y) = 0+0.19$

$E(y) = 0.19$

Variance is calculated as:

$Var(y) = E(y^2) - (E(y))^2$

Calculate $E(y^2)$

$E(y^2) = \sum(y^2 * P(y))$

$E(y^2) = 0^2 * 0.81 + 1^2 * 0.19$

$E(y^2) = 0 + 0.19$

$E(y^2) = 0.19$

So:

$Var(y) = E(y^2) - (E(y))^2$

$Var(y) = 0.19 - 0.19^2$

$Var(y) = 0.19 - 0.0361$

$Var(y) = 0.1539$

Solving (e): The covariance and the coefficient of correlation

Covariance is calculated as:

$COV(x,y) = E(xy) - E(x) * E(y)$

Calculate E(xy)

$E(xy) = \sum (xy * P(xy))$

This gives:

$E(xy) = x_0y_0 * P(0,0) + x_1y_0 * P(1,0) +x_0y_1 * P(0,1) + x_1y_1 * P(1,1)$

$E(xy) = 0*0 * 0.63 + 1*0 * 0.18 +0*1 * 0.09 + 1*1 * 0.1$

$E(xy) = 0+0+0 + 0.1$

$E(xy) = 0.1$

So:

$COV(x,y) = E(xy) - E(x) * E(y)$

$Cov(x,y) = 0.1 - 0.28 * 0.19$

$Cov(x,y) = 0.1 - 0.0532$

$Cov(x,y) = 0.0468$

The coefficient of correlation is then calculated as:

$r = \frac{Cov(x,y)}{\sqrt{Var(x) * Var(y)}}$

$r = \frac{0.0468}{\sqrt{0.2016 * 0.1539}}$

$r = \frac{0.0468}{\sqrt{0.03102624}}$

$r = \frac{0.0468}{0.17614266944}$

$r = 0.26569371378$

$r \approx 0.2657$ --- approximated

8 0

3 years ago