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STatiana [176]
3 years ago
8

10 m 8 m 6.2 m 8 m What’s the area

Mathematics
1 answer:
hjlf3 years ago
8 0

Answer:

what's the shape? that's important to

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A botanist measures the heights of 16 seedlings and obtains a mean and standard deviation of 72.5 cm and 4.5 cm, respectively. F
Brums [2.3K]

Answer:

c. [70.30, 74.70]

Step-by-step explanation:

Confidence interval is a range of values in which there is a specified probability that the value of a parameter lies within that range.

The confidence interval of a statistical data can be written as.

x+/-zr/√n .......1

Given;

Mean gain x = 72.5 cm

Standard deviation r = 4.5 cm

Number of samples n = 16

Confidence interval = 95%

z (at 95% confidence) = 1.96 (from the table)

Substituting the values we have into equation 1;

72.5 +/- 1.96×4.5/√16

72.5 +/- 2.20 cm

[70.30, 74.70]

Therefore, the 95% confidence interval is

[70.30, 74.70] cm

4 0
3 years ago
Using the 28/36 ratio, determine the maximum allowable recurring debt for someone with a monthly income of $3,200. A. $256 b. $5
stiv31 [10]
The answer is a. $256

Step-by-step explanation:

The computation of the maximum allowable recurring debt is shown below:

Given ratio = 28:36

And the monthly income is $3,200

So the maximum expense on housing = 28% of $3,200 = $896

And, the maximum expense on total debt = 36% of $3,200 = $1,152

Now the maximum alloweable recurring debt is

= $1,152 - $896

= $256
6 0
3 years ago
Help me on this math problem?
shusha [124]

Answer:

just add them

Step-by-step explanation:

4 0
3 years ago
Joel can run around a 14– mi track in 81 sec, and Jason can run around the track in 69 sec. If the runners start at the same poi
Alenkasestr [34]

Answer:37.2 sec

Step-by-step explanation:

Given

Joel takes 81 sec for 14 mile long track

speed of joel \frac{14}{1.35} mi/min=10.37 mi/min

Jason takes 69 sec for 14 mi track

so jason speed is \frac{14}{1.15} mi/min=12.17 mi/min

if both starts from same spot in opposite direction then

\frac{x}{10.37}=\frac{14-x}{12.17}

where x is distance covered by Joel

then x=6.44 miles

therefore time required

t=\frac{6.44}{10.37}=0.621 min\approx 37.2 sec

8 0
4 years ago
PLZZZ I NEEDDD HELLPP ITS Give the place value of the number five in each of the numbers below HELP AAAAAA
musickatia [10]
50,000 okkkkklike yeahh
3 0
4 years ago
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