How do you do <br> -5(3m+6)=-3(4m-2)

if the ratio to cats to dogs at the city pound is 92 to 144. which statement describes this relationship?

nadya68 [22]

Answer:

For every 92 cats in the city pound, there are 144 dogs.

OR

For every 144 dogs in the city pound, there are 92 cats

Step-by-step explanation:

You didn't provide the options, so that was my best shot at it.

7 0

2 years ago

Read 2 more answers

(5x+3y=-3<br> | 3x – 3y =-21

disa [49]

Answer:

8x = -24 => x = -3

y = 4

Step-by-step explanation:

7 0

3 years ago

Solve for the variables

Vesna [10]

Answer:

P = 32

x = 35

Step-by-step explanation:

For the first diagram,

The two angles given are corresponding angles. Corresponding angles are equal;

So;

the first angle is a right angle triangle;

90 = 3P - 6

3P = 90 + 6

3P = 96

P = 32

For the second diagram;

The sum of those two angles is 180°

(x + 75) + 2x = 180

x + 2x + 75 = 180

3x + 75 = 180

3x = 180 - 75

3x = 105

x = 35

5 0

3 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 $p_2$ .