Answer:
it is y = 2x + 3
Step-by-step explanation:
i just took the test
Answer
Given,
Mass of bag = 6 Kg
Area covered by the bag = 2.5 x 5 = 12.5 m²
a) Bags required to cover area = 7.5 x 10.2
= 76.5 m²
Number of bags, N = 
N = 6.12 = 7 (approx.)
7 Bags of 6 Kg fertilizers will be needed to cover the area.
b) Cost of 6 Kg fertilizers = $15.50
Cost of 7 bags of 6 Kg fertilizers = 7 x $15.50
= $108.50.
Answer:
The equation are:
1) h + s = 20
2) 4h = 1s
Step-by-step explanation:
Number of hats Elliott made = h
Number of scarfs Elliott made = s
total number of items Elliott made = 20
h + s = 20 ..[1]
1 hat uses 0.2 kilograms of yarn
1 hat = 0.2 kg of yarn
Then h hats will use = 0.2h kg of yarn
1 scarf uses 0.1 kilograms of yarn
1 scarf= 0.1 kg of yarn
Then s scarfs will use = 0.1s kg of yarn
She wants to twice as much as yarn for scarves for hats:
2 × (0.2h kg of yarn) = 0.1s kg of yarn
0.4h = 0.1s
4h = 1s...[2]
The equation are:
1) h + s = 20
2) 4h = 1s
Answer:
Here we have given two catogaries as degree holder and non degree holder.
So here we have to test the hypothesis that,
H0 : p1 = p2 Vs H1 : p1 not= p2
where p1 is population proportion of degree holder.
p2 is population proportion of non degree holder.
Assume alpha = level of significance = 5% = 0.05
The test is two tailed.
Here test statistic follows standard normal distribution.
The test statistic is,
Z = (p1^ - p2^) / SE
where SE = sqrt[(p^*q^)/n1 + (p^*q^)/n2]
p1^ = x1/n1
p2^ = x2/n2
p^ = (x1+x2) / (n1+n2)
This we can done in TI_83 calculator.
steps :
STAT --> TESTS --> 6:2-PropZTest --> ENTER --> Input all the values --> select alternative "not= P2" --> ENTER --> Calculate --> ENTER
Test statistic Z = 1.60
P-value = 0.1090
P-value > alpha
Fail to reject H0 or accept H0 at 5% level of significance.
Conclusion : There is not sufficient evidence to say that the percent of correct answers is significantly different between degree holders and non-degree holders.
