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2 years ago

What the area for thos? A= 1/2 (bxh)

Mathematics

1 answer:

lidiya [134]2 years ago

4 0

Answer:

A =8.75 ft^2

Step-by-step explanation:

The area of the triangle is

A = 1/2 bh where b is the base and h is the height

A = 1/2 ( 7 * 2.5)

A =8.75 ft^2

You might be interested in

Find the length of side x in simplest radical form with a rational denominator

Angelina_Jolie [31]

Answer:

The length of x=3/1

Step-by-step explanation:

Given the right triangle

Ф = 30

We know that

sin Ф = opposite/hypotenuse

The opposite of angle 30° = x

The hypotenuse is = 6

sin 30 = opposite/hypotenuse

1/2 = x/6 ∵ sin 30 = 1/2

1/2 × 6 = x

6/2 = x

3/1 = x

Therefore, the length of x=3/1

4 0

3 years ago

If it takes 5 years for an animal population to double, how many years will it take until the population

muminat

9514 1404 393

Answer:

7.92 years

Step-by-step explanation:

We want to find t such that ...

3 = 2^(t/5)

where 2^(t/5) is the annual multiplier when doubling time is 5 years.

Taking logs, we have ...

log(3) = (t/5)log(2)

t = 5·log(3)/log(2) ≈ 7.92 . . . years

It will take about 7.92 years for the population to triple.

6 0

2 years ago

If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal

lorasvet [3.4K]

Answer:

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

The probability is $P(Z > 2.5 ) = 0.0062097$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 800$

The variance is $var(x) = 1600 \ kg$

The range consider is $x_1 = 778 \ kg \ x_2 = 834 \ kg$

The value consider in second question is $x = 900 \ kg$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{var (x)}$

substituting value

$\sigma = \sqrt{1600}$

$\sigma = 40$

The percentage of a cucumber give the crop amount between 778 and 834 kg is mathematically represented as

$P(x_1 < X < x_2 ) = P( \frac{x_1 - \mu }{\sigma} < \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma } )$

Generally $\frac{X - \mu }{ \sigma } = Z (standardized \ value \ of \ X)$

$P(x_1 < X < x_2 ) = P( \frac{778 - 800 }{40} < Z< \frac{834 - 800 }{40 } )$

$P(x_1 < X < x_2 ) = P(z_2 < 0.85) - P(z_1 < -0.55)$

From the z-table the value for $P(z_1 < 0.85) = 0.80234$

and $P(z_1 < -0.55) = 0.29116$

$P(x_1 < X < x_2 ) = 0.80234 - 0.29116$

$P(x_1 < X < x_2 ) = 0.51118$

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

The probability of cucumber give the crop exceed 900 kg is mathematically represented as

$P(X > x ) = P(\frac{X - \mu }{\sigma } > \frac{x - \mu }{\sigma } )$

substituting values

$P(X > x ) = P( \frac{X - \mu }{\sigma } >\frac{900 - 800 }{40 } )$

$P(X > x ) = P(Z >2.5 )$

From the z-table the value for $P(Z > 2.5 ) = 0.0062097$

7 0

3 years ago

Solve for q. q/18=5 q=

Nataliya [291]

Answer:

q =90

Step-by-step explanation:

q/18=5

Multiply each side by 18

q/18 * 18=5* 18

q =90

3 0

2 years ago

Read 2 more answers

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection = </em>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago