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givi [52]
3 years ago
6

Find the area : (see question in picture)

Mathematics
1 answer:
quester [9]3 years ago
4 0

I cant really see the full question

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Reduce 105 in the ratio 3:5​
Komok [63]

9514 1404 393

Answer:

  39.375 : 65.625

Step-by-step explanation:

The meaning of "reduce" in this context is unclear. We assume you want to divide the quantity 105 into parts that have a ratio of 3 : 5. Then the smaller part will be 3/(3+5) = 3/8 of the total; the larger part will be 5/8 of the total.

  3/8 × 105 = 39 3/8

  5/8 × 105 = 65 5/8

The divided quantities are ...

  39.375 : 65.625

3 0
3 years ago
3 7/8p+ 2 2/3p please i need help!!
SIZIF [17.4K]
Did you figure it out
8 0
3 years ago
The equation for the speed/distance
asambeis [7]

Step-by-step explanation:

S = 93 log (d) + 65; S = wind speed; d = distance traveled by tornado

A) tornado travels 130 miles = d; find S

S = 93 log10(130) + 65

S = 261.6 mph rounds up to 262 mph

7 0
3 years ago
Read 2 more answers
If 1 1/4 kg sugar cost 2.25, find the cost of 3 1/4 kg of sugar
GalinKa [24]
<h2><u><em>$6.25</em></u></h2><h2><u><em></em></u></h2>

Step-by-step explanation:

because 1 kg is worth 2 is 1/4 is worth .25

so multiply 2 by 3 and add 1/4

good luck hope this helps.

3 0
3 years ago
Read 2 more answers
he average movie admission price for a recent year was $7.18. The population variance was 3.81. A random sample of 15 theater ad
Gemiola [76]

Answer:

Since the calculated value of  z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.

Step-by-step explanation:

The data given is

Population mean μ= $ 7.18

Population variance= σ²= 3.81

Population Standard Deviation = √σ²= √3.81= 1.952

Sample Mean= x`= $ 8.02

Sample Standard Deviation =s = $ 2.08

Sample Size = 15

Significance Level = ∝= 0.05

The null and alternate hypotheses are

H0: σ1=σ2 against the claim that Ha: σ1≠ σ2

where  σ1 is the population variance and

σ2 is the sample variance

The rejection region is Z ≥ ±1.96 for two tailed test  at ∝= 0.05

The test statistic z is used

z= x`-  μ/ σ/√n

Putting the values

Z= 8.02-7.18/1.952/√15

z= 0.84/0.51599

z= 1.6279

Since the calculated value of  z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.

3 0
3 years ago
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