All categories

2 years ago

Please help my finale math exam is do to night

Mathematics

1 answer:

Misha Larkins [42]2 years ago

4 0

Answer:

Dimension volume

6 cups (1.5 L) 8 x 2 inch (20 x 5 cm) Round Cake Pan

6 cups (1.5 L) 9 x 1.5 inch (23 x 4 cm) Round Cake Pan

6 cups (1.5 L) 8 x 8 x 1.5 inch (20 x 20 x 4 cm) Square Cake Pan

8 cups (2 L) 9 x 2 inch (23 x 5 cm) Round Cake Pan

here are four outcomes that sum to seven out of 24 possible outcomes. Therefore the probability of values of four sided die and six sided

Step-by-step explanation:

You might be interested in

Plz help me with the ones that are circled

Alisiya [41]

Your answer for your question is going to be.... um.... sorry I forgot

3 0

3 years ago

30 points!

Andrei [34K]

Its the first one. Beacuse she gains speed then slows down and stops but then starts up again then slows down. Hope this helps!

7 0

3 years ago

Read 2 more answers

What is 3 and 8 tenths as a decimal

BartSMP [9]

3.8 is it as a decimal

6 0

3 years ago

Danny bought 5 Pes to $0.30 each six pencils at $0.25 since in The Notebook for $3 and the tax is $8 how much did he pay will ma

Paha777 [63]

The answer is 14. This is because 0.30 multiplied by 5 is 1.5. Then 6 multiplied by 0.25 is 1.5 as well. Adding these two together is 3. 3 plus 3 for the notebook is 6. Than 6 plus 8 dollar tax is 14 dollars.

6 0

3 years ago

Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean

Zinaida [17]

Answer:

$E[R]$ = 99 Ω

$\sigma_R$ = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, $E[R] = \frac{a+b}{2} = \frac{95+103}{2} = \frac{198}{2}$ = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

$V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333$

Hence,

The standard deviation of resistance, $\sigma_R = \sqrt{V(R)} = \sqrt5.333$ = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

$z_1 = \frac{102-99}{2.3094} = 1.299$

$z_2 = \frac{98-99}{2.3094} = -0.433$

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

5 0

3 years ago