Pls help What is the value of t? A. 150 B. 60 C. cannot be determined D. 30

Can someone convert 5cm to milimeter?

zhenek [66]

Answer: 50 millimeters

Step-by-step explanation:

5 cm is one 100th of a meter. Millimeters are one 1000th. Therefore, 1 of each unit in cm is 10 units in mm

6 0

2 years ago

Read 2 more answers

Find values a and b that satisfy a∙〈5,-4,0〉+b∙〈2,2,-3〉=〈11,-16,-6〉.

WARRIOR [948]

The right hand side of the equation is 〈11 , -16 , 6〉

The values of a and b are a = 3 , b = -2

Step-by-step explanation:

If n . <a , b> + m . <c , d> = < x , y>, then

na + mc = x
nb + md = y

∵ a ∙〈5 , -4 , 0〉 + b ∙ 〈2 , 2 , -3〉 = 〈11 , -16 , 6〉

- By using the rule above

∴ a(5) + b(2) = 11 ⇒ (1)

∴ a(-4) + b(2) = -16 ⇒ (2)

∴ a(0) + b(-3) = 6 ⇒ (3)

Use equation (3) to find the value of b

∵ a(0) + b(-3) = 6

∴ 0 - 3b = 6

∴ -3b = 6

- Divide both sides by -3

∴ b = -2

Substitute b in equation (1) or equation (2) to find a

∵ a(5) + b(-2) = 11

∴ 5a - 2b = 11

∵ b = 2

∴ 5a - 2(2) = 11

∴ 5a - 4 = 11

- Add 4 to both sides

∴ 5a = 15

- Divide both sides by 5

∴ a = 3

To check your answer substitute a and b in equation (2)

∵ a(-4) + b(2) = -16

∴ -4a + 2b = -16

∵ a = 3 , b = -2

∵ The left hand side is -4a + 2b = -4(3) + 2(-2) = -12 - 4 = -16

∵ The right hand side is -16

∴ L.H.S = R.H.S

∴ The values of a and b are 3 , -2

The values of a and b are a = 3 , b = -2

Learn more:

You can learn more about the equations in brainly.com/question/11306893

#LearnwithBrainly

8 0

3 years ago

A population of 30 deer is introduced into a wildlife sanctuary. It is estimated that the sanctuary can sustain up to 200 deer,

gogolik [260]

Answer:

After one year the population will be 33 deers, and after two years it will be 36 deers.

Step-by-step explanation:

Given that a population of 30 deer is introduced into a wildlife sanctuary, and it is estimated that the sanctuary can sustain up to 200 deer, and absent constraints, the population would grow by 10% per year, to predict the population after one year and after two years, the following calculations must be performed:

A)

30 x 1.1 = X

33 = X

B)

30 x 1.1 ^ 2 = X

30 x 1.21 = X

36.3 = X

Therefore, after one year the population will be 33 deers, and after two years it will be 36 deers.

7 0

2 years ago

50 students live in a dormitory. The parking lot has the capacity for 30 cars. Each student has a car with probability 12 , inde

Maurinko [17]

Answer:

P(X is greater than 30) = 0.06

Step-by-step explanation:

Given that:

Sample proportion (p) = 0.5

Sample size = 30

The Binomial can be approximated to normal with:

$\mu = np = 50 \times 0.5 \\ \\ \mu= 25$

$\sigma = \sqrt{np(1-p) } \\ \\ \sigma = \sqrt{50 \times (0.5)(1-0.5) } \\ \\ \sigma = 3.536$

To find:

P(X> 30)

So far we are approximating a discrete Binomial distribution using the continuous normal distribution. 30 lies between 29.5 and 30.5

Normal distribution:

x = 30.5, $\mu$ = 25, $\sigma$ = 3.536

Using the z test statistics;

$z = \dfrac{x - \mu}{\sigma}$

$z = \dfrac{30.5 - 25}{3.536}$

$z = \dfrac{5.5}{3.536}$

z = 1.555

The p-value for P(X>30) = P(Z > 1.555)

The p-value for P(X>30) = 1 - P (Z< 1.555)

From the z tables;

P(X> 30) = 1 - 0.9400

Thus;

P(X is greater than 30) = 0.06

7 0

2 years ago

Element X decays radioactively with a half life of 12 minutes. If there are 200

PolarNik [594]

$\textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{current amount}\dotfill &20\\ P=\textit{initial amount}\dotfill &200\\ t=\textit{elapsed time}\\ h=\textit{half-life}\dotfill &12 \end{cases} \\\\\\ 20=200\left( \frac{1}{2} \right)^{\frac{t}{12}}\implies \cfrac{20}{200}=\left( \frac{1}{2} \right)^{\frac{t}{12}}\implies \cfrac{1}{10}=\left( \frac{1}{2} \right)^{\frac{t}{12}}$

$\log\left( \cfrac{1}{10} \right)=\log\left[ \left( \frac{1}{2} \right)^{\frac{t}{12}} \right]\implies \log\left( \cfrac{1}{10} \right)=t\log\left[ \left( \sqrt[12]{\frac{1}{2}} \right) \right] \\\\\\ \cfrac{\log\left( \frac{1}{10} \right)}{\log\left[ \left( \sqrt[12]{\frac{1}{2}} \right) \right]}=t\implies \implies \stackrel{mins}{39.9}\approx t$

8 0

2 years ago