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1 year ago

Please help I will give brainliest.

Mathematics

1 answer:

Kruka [31]1 year ago

8 0

Answer:

Step-by-step explanation:

Construct a perpendicular bisector of segment AB.

Perpendicular bisector is a line that divides the line into two equal halves forming 90° at the intersection point.

So, points A and B will be at the same distance from the intersection point.

If the constructed line (perpendicular bisector) is the reflection axis, then point B will be the refection point of A.

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Daniel had $25 to spend at the fair if the admission to the fair was four dollars in the rides cost $1.50 each what is the great

Sever21 [200]

Answer:

Step-by-step explanation:

25-4=21

21/1.5=14

6 0

3 years ago

Read 2 more answers

Can someone help me with this math homework please!

mamaluj [8]

Answer:

The rock hits the ground between <u>2</u> seconds and <u>2.5</u> seconds after it is dropped

Step-by-step explanation:

The given table is presented as follows;

$\begin{array}{ccl}t&h(t)&Description\\0&20&Initial \ height\\0.5&18.8&Rock \ in \ downward \ motion\\1&15.1&\\1.5&9&\\2&0.4&The \ height \ just \ before \ the \ rock \ hits \ the \ ground \\2.5&-10.6&The \ calculated \ height \ after\ the \ rock \ hits \ the \ ground \\3&-24.1&Calculated \ height \ after\ the \ rock \ hits \ the \ ground\end{array}$

Therefore, the rock hits the ground between t = 2 seconds and t = 2.5 seconds after it is dropped.

4 0

3 years ago

Forty percent of households say they would feel secure if they had $50,000 in savings. you randomly select 8 households and ask

Kay [80]

Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

$P(x=r)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

If 8 households choose randomly,

That is, n = 8

(a) the probability of the number that say they would feel secure is exactly 5

$P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}$

$=56(0.4)^5 (0.6)^3$

$=0.12386304$

(b) the probability of the number that say they would feel secure is more than five

$P(X>5) = P(X=6)+ P(X=7) + P(X=8)$

$=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}$

$=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8$

$=0.04980736$

(c) the probability of the number that say they would feel secure is at most five

$P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$

$=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}$