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Mrrafil [7]
3 years ago
6

Jeannie is 5 years older than her younger sister April. Together, Jeannie and April’s ages add to 41. How old is April?

Mathematics
1 answer:
Rzqust [24]3 years ago
6 0
Apirl is 18 years old and her older sister is 23

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4. Joe received an hourly wage of $8.15. His boss gave him a 7% raise. How much does Joe make per hour now?
Olenka [21]

Answer:

$ 8.72

Step-by-step explanation:

  • How much does Joe make per hour now?

$8.15 + 7%·$8.15 =

= $8.15 + 7×8.15/100

= $8.15 + $57.05/100

= $8.15 + $0.57

= $ 8.72

5 0
3 years ago
Read 2 more answers
Four sets of data are shown in box-and-whisker plots. Which set has the largest 3rd Quartile value?
Airida [17]

Answer:

I think that is A.

Step-by-step explanation:

7 0
3 years ago
Round 3308.89 to the nearest australian dollar
const2013 [10]
Round 3308.89 to the nearest Australian dollar - $4336 AUD .
3 0
3 years ago
The volumes of two cylindrical cans of the same shape vary directly as the cubes of their radii. If a can with a six-inch radius
Cerrena [4.2K]

Answer:

The second similar can with a radius of 24-inch radius will hold 12 gallons

Step-by-step explanation:

The given parameters are;

The volumes of the cylindrical can ∝ (The radius of the cans)³

The volume of first can, V₁ = 1¹/₂ pints

The radius of the first can, r₁ = 6-inch

The radius of the second can, r₂ = 24-inch

Therefore, we have;

V ∝ r³

V = k × r³

∴ 1¹/₂ pints ∝ (6 in.)³

1¹/₂ pints = 43.3125 in³

∴ 43.3125 in³ = k × 216 in.³

The constant of proportionality, k = 43.3125/216 = 0.2005208333

Therefore, we have for the second can, we have;

V₂ = k × r₂³ = 43.3125/216 × (24 in.)³ = 2772 in.³

The volume of the second can = 2772 in.³

1 in.³ = 0.004329 gallons

∴ 2772 in.³ = 2772 × 0.004329 gallons = 12 gallons

Therefore, the volume the second similar can with a 24-inch radius will hold = 12 gallons.

8 0
3 years ago
To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil
Debora [2.8K]

Answer:

statistic value t= 5.43

p-value: < 0.00001.

Step-by-step explanation:

Hello!

To obtain information over the corrosion-resistance properties of a certain type of steel conduit a random sample of 45 specimens was taken and buried for two years.

The study variable is:

X: Max. penetration of a steel conduit.

The data of the sample

n= 45

sample mean X[bar]= 53.4

sample standard deviation S= 4.2

The conduits are manufactured to have a true average penetration of at most 50 mills, symbolically: μ ≤ 50

The hypothesis is:

H₀: μ ≤ 50

H₁: μ > 50

α: 0.05

To choose the corresponding statistic to use to study the population mean, the variable must have a normal distribution. There is no available information to check this, so I'll just assume that the variable has a normal distribution and, since the population variance is unknown and the sample is small, the statistic to use is a Student t.

Under the null hypothesis, the critical region and the p-value are one-tailed.

Critical value:

t_{n-1; 1-\alpha } = t_{44; 0.95} = 1.68

Rejection rule:

Reject the null hypothesis when t ≥ 1.68

t= \frac{53.4 - 50}{\frac{4.2}{\sqrt{45} } }

t= 5.43

The calculated value is greater than the critical value, thedecision is to rject the null hypothesis.

p-value:

P(t ≥ 5.43) = 1 - P(t < 5.43) = < 0.00001.

The p-value is less than α so the decision is to reject the null hypothesis.

Since the null hypothesis was rejected, then the population average of the penetration of the conduits specimens is greater than 50 mils. It is not recommendable to use these conduits.

I hope it helps!

4 0
3 years ago
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