Answer:
<u>Answer (a):</u>
s + l = 22 ... (i)
43s + 75l = 1234 ... (ii)
<u>Answer (b)</u>
9 large dogs.
Step-by-step explanation:
Paws at Play made a total of $1234 grooming 22 dogs.
Paws at Play charges $43 to groom each small dog and;
$75 for each large dog.
Let the number of small dogs be 's'
And the number of large dogs be 'l'
<u>A system of equations will be:</u>
s + l = 22 ... (i)
43s + 75l = 1234 ... (ii)
Solving this set of simultaneous equations by elimination, we simply multiply (i) by 43 to get;
43s + 43l = 946 ... (i)
43s + 75l = 1234 ... (ii)
Subtracting (i) from (ii) we get;
32l = 288 , l = 
= 9
So there are 9 large dogs.
Answer:
240 
12.5p + 65
14 
p
Step-by-step explanation:
I'll explain the equation in different parts
12.5p - Since the number of times he visits is unknown, it has to be a variable. With each visit, however, he earns 12.5 points.
65 - This is a set value that remains constant
240 - Since he needs AT LEAST 240 points, he needs 240 points or more to get his free ticket
As for solving the equation just use properties of equality
240 12.5p + 65
175 12.5p
14 p
Answer:
1.16
Step-by-step explanation:
Given that;
For some positive value of Z, the probability that a standardized normal variable is between 0 and Z is 0.3770.
This implies that:
P(0<Z<z) = 0.3770
P(Z < z)-P(Z < 0) = 0.3770
P(Z < z) = 0.3770 + P(Z < 0)
From the standard normal tables , P(Z < 0) =0.5
P(Z < z) = 0.3770 + 0.5
P(Z < z) = 0.877
SO to determine the value of z for which it is equal to 0.877, we look at the
table of standard normal distribution and locate the probability value of 0.8770. we advance to the left until the first column is reached, we see that the value was 1.1. similarly, we did the same in the upward direction until the top row is reached, the value was 0.06. The intersection of the row and column values gives the area to the two tail of z. (i.e 1.1 + 0.06 =1.16)
therefore, P(Z ≤ 1.16 ) = 0.877
Answer:
Step-by-step explanation:
Since population std deviation is not known, we use t critical value for testing of hypothesis.
a) 90% conf level when n =28
df = 27, t critical = 1.706
b) 95% 28
df =27 and t critical = 2.052
c) 90% and 15
df =14 and t critical = 1.761
d) 95% and n =15
df =14: t critical = 2.145
Answer:
B. 42-10
Step-by-step explanation:
It is the amount that Sid has minus what he ows
