The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish’s mouth. suppose the archerfish squirts water with an initial speed of 2.30 m/s at an angle of 19.5° above the horizontal. when the stream of water reaches a beetle on a leaf at height h above the water’s surface it is moving horizontally. (a) how much time does the beetle have to react? (b) what is the height ℎ of the beetle? (c) what is the horizontal distance d between the fish and the beetle when the water is launched?

thats, yes

Step-by-step explanation:

i think

*The dot plot is shown in the attachment below

2

Step-by-step explanation:

Interquartile range is the difference between the upper median (Q3) and the lower median (Q1).

First, let's write out each value given in the data. Each dot represents a data point.

We have:

2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7

=>Find the median:

Our median is the middle value. The middle value is the 6th value = 4

==>Upper median Q3) = the middle value of the set of data we have from the median to our far right.

2, 3, 3, 4, 4, |4,| 4, 5, [5], 6, 7

Our upper median = 5

==>Lower median(Q1) = the middle value of the data set we have from our median to our far left.

2, 3, [3], 4, 4, |4,| 4, 5, 5, 6, 7

Lower median = 3

==>Interquartile range = Q3 - Q1 = 5-3 = 2