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3 years ago

6

driving to and from the butterfly garden took 45 minutes each way. Students spent 3 hours in the garden and 30 minutes eating lu

nch. If the groups left school at 9:00a.m. what time did they get back

1 answer:

Setler [38]3 years ago

3 0

45 min add 45 min equals 1:30 add 3 3:30 add 30 equals they got vack at 1:00pm

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Simplify 2(12-18\3+4)

Goshia [24]

${\bold{\red{\huge{\mathbb{QUESTION}}}}}$

Simplify

$2(12 - \frac{18}{3} + 4)$

$\huge\green{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}}}$

ATQ->

$2(12 - ( \frac{ \cancel{18}^{ \: \: \: 6} }{3} ) + 4 \\ 2(12 - 6 + 4) \\ 2(16 - 6) \\ 2(10) \\ 20 \: \: is \: your \: ans$

4 0

2 years ago

Read 2 more answers

Please answer this<br>Hope you get along<br>I report What to answer

Llana [10]

What the hel you expect us to do all that for 5 kids points? Go ask you mom to do that for you. If you're asking US to do that, then it has to be 40 or 50 points.

3 0

2 years ago

Find the indicated probability.

natka813 [3]

Answer:

C) 0.179

Step-by-step explanation:

Since the trials are independent, this is a binomial distribution:

<u>Recall:</u>

Binomial Distribution --> $P(k)={n\choose k}p^kq^{n-k}$
$P(k)$ denotes the probability of $k$ successes in $n$ independent trials
$p^k$ denotes the probability of success on each of $k$ trials
$q^{n-k}$ denotes the probability of failure on the remaining $n-k$ trials
${n\choose k}=\frac{n!}{(n-k)!k!}$ denotes all possible ways to choose $k$ things out of $n$ things

<u>Given:</u>

$n=10$
$k=4$
$p^k=0.53^4$
$q^{n-k}=(1-0.53)^{10-4}=0.47^6$
${n\choose k}={10\choose 4}=\frac{10!}{(10-4)!4!}=210$

<u>Calculate:</u>

$P(4)=(210)(0.53^4)(0.47^6)=0.1786117069\approx0.179$

Therefore, the probability that the archer will get exactly 4 bull's-eyes with 10 arrows in any order is 0.179

7 0

1 year ago

The equation a=1/2(b^1+b^2)h can be determined the area, a, of a trapezoid with height, h, and base lengths, b^1 and b^2 Which a

Evgesh-ka [11]

The complete question is as follows.

The equation a = $\frac{1}{2}(b_1 + b_2 )h$ can be used to determine the area , <em>a</em>, of a trapezoid with height , h, and base lengths, $b_1$ and $b_2$ . Which are equivalent equations?

(a) $\frac{2a}{h} - b_2 = b_1$

(b) $\frac{a}{2h} - b_2 = b_1$

(c) $\frac{2a - b_2}{h}$ = $b_1$

(d) $\frac{2a}{b_1 + b_2} = h$

(e) $\frac{a}{2(b_1 + b_2)}$ = h

Answer: (a) $\frac{2a}{h} - b_2 = b_1$ ; (d) $\frac{2a}{b_1 + b_2} = h$ ;

Step-by-step explanation: To determine $b_1$ :

a = $\frac{1}{2}(b_1 + b_2 )h$

2a = ( $b_1 + b_2$ )h

$\frac{2a}{h} = b_1 + b_2$

$\frac{2a}{h} - b_2 = b_1$

To determine h:

a = $\frac{1}{2}(b_1 + b_2 )h$

2a = $(b_1 + b_2)h$

$\frac{2a}{(b_1 + b_2)}$ = h

To determine $b_2$

a = $\frac{1}{2}(b_1 + b_2 )h$

2a = $(b_1 + b_2)h$

$\frac{2a}{h} = (b_1 + b_2)$

$\frac{2a}{h} - b_1 = b_2$

Checking the alternatives, you have that $\frac{2a}{h} - b_2 = b_1$ and $\frac{2a}{(b_1 + b_2)}$ = h, so alternatives <u>A</u> and <u>D</u> are correct.

4 0

3 years ago

1.There are 32 adults and 20 children at a shool play.Which of these ratios is not equivalent to the ratio of adults to children

OLEGan [10]

Well it have to be D because 32 and 20 can not go into 64 to 40 because 32 times 64 is 2048 and 64 times 20 is 1280 and if it was 40 times 32 and 20 it will still be higher. I HAVE FAITH IN YOU.

7 0

2 years ago