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sergij07 [2.7K]

3 years ago

14

A group of students went on a field trip. The students were divided into 6 buses and each bus had less than 52 students. How man

y students went on the field trip?

2 answers:

Vikki [24]3 years ago

7 0

Answer:

A maximum of 306

Step-by-step explanation:

You can estimate that since each bus had less than 52 students, that 51 were on each bus. 51 times 6=306

lozanna [386]3 years ago

7 0

51x6=306
Up to 306 students in a bus

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A running circuit is in the shape of a triangle with lengths of 6km, 6.5km and 7km. What are the sizes of the angles (in minutes

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A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to $180^{o}$ . The <em>measures</em> of the internal <u>angles</u> of the <u>triangle</u> given in the question are A = $52.6^{o}$ , B = $59.4^{o}$ , and C = $68^{o}$ .

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to $180^{o}$ .

Considering the given question, let the <u>sides</u> of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the <em>Cosine rule</em> to have:

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab Cos C

So that;

$7^{2}$ = $6^{2}$ + $(6.5)^{2}$ - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

= 29.25

Cos C = $\frac{29.25}{78}$

= 0.375

C = $Cos^{-1}$ 0.375

= 67.9757

C = $68^{o}$

Apply the <em>Sine rule</em> to determine the <u>value</u> of B,

$\frac{b}{Sin B}$ = $\frac{c}{Sin C}$

$\frac{6.5}{Sin B}$ = $\frac{7}{Sin 68}$

SIn B = $\frac{6.5 *Sin 68}{7}$

= 0.861

B = $Sin^{-1}$ 0.861

= 59.43

B = $59.4^{o}$

Thus to determine the value of A, we have;

A + B + C = $180^{o}$

A + $59.4^{o}$ + $68^{o}$ = $180^{o}$

A = $180^{o}$ - 127.4

= 52.6

A = $52.6^{o}$

Therefore the <u>sizes</u> of the <em>internal angles</em> of the triangle are: A = $52.6^{o}$ , B = $59.4^{o}$ , and C = $68^{o}$ .

For more clarifications on applications of the Sine and Cosine rules, visit: brainly.com/question/14660814

#SPJ1

8 0

2 years ago

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Nuetrik [128]

Answer:

3.70% probability she allows more than one goal

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

X~Pois(0.3)

This means that $\mu = 0.3$

In her next match, what is the probability she allows more than one goal?

Either she allows at most one goal, or she allows more than one goal. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 1) + P(X > 1) = 1$

We want $P(X > 1)$ . So

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.3}*(0.3)^{0}}{(0)!} = 0.7408$

$P(X = 1) = \frac{e^{-0.3}*(0.3)^{1}}{(1)!} = 0.2222$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.7408 + 0.2222 = 0.9630$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.9630 = 0.0370$

3.70% probability she allows more than one goal

3 0

3 years ago