Please help I give 20 points please no fake answers =/

The admission fee at an amusement park is $4.00 for children and $5.60 for adults. On a certain day, 284 people entered the park

miv72 [106K]

4c+5.6d=1360
Let
Children (c)=x
Adults (d)=284-x

4x+5.6 (284-x)=1360
Solve for x
4x+1590.4-5.6x=1360
4x-5.6x=1360-1590.4
-1.6x=-230.4
X=230.4÷1.6
X=144 children

284−144=140 adults

3 0

3 years ago

What is the most precise name for quadrilateral ABCD with vertices A(–5,2), B(–3, 5),C(4, 5),and D(2, 2)?

bonufazy [111]

Answer: ABCD is a parallelogram.

Step-by-step explanation:

First we plot these point on a graph as given in attachment.

From the attachment we can observe that AD || BC || x-axis .

also, AB ||CD, that will make ABCD a parallelogram , but to confirm we check the property of parallelogram "diagonals bisect each other" , i.e . "Mid point of both diagonals are equal".

Mid point of AC= $(\dfrac{-5+4}{2},\dfrac{2+5}{2})=(\dfrac{-1}{2},\dfrac{7}{2})$

Mid point of BD= $(\dfrac{-3+2}{2},\dfrac{5+2}{2})=(\dfrac{-1}{2},\dfrac{7}{2})$

Thus, Mid point of AC=Mid point of BD

i.e. diagonals bisect each other.

That means ABCD is a parallelogram.

8 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .