All categories

3 years ago

Name two different numbers that round to 3.8 when rounded to the nearest tenth.

Mathematics

2 answers:

AveGali [126]3 years ago

6 0

Answer:

3.78 and 3.79

You aren't limited to just one decimal place.

maxonik [38]3 years ago

3 0

Two numbers that round to 3.8 when rounding to the nearest tenth can be 3.78 and 3.79

You might be interested in

Half of a pizza was broccoli and half was mushroom. george ate 1/3 of the broccoli part and 1/4 of the mushroom part. how much o

butalik [34]

1/3x1/4=1/16 I think

5 0

4 years ago

A researcher wishes to determine whether people with high blood pressure can lower their blood pressure by performing yoga exerc

Anastaziya [24]

Answer:

90% confidence interval for the difference between the two population means

( -23.4166 , -6.5834)

Step-by-step explanation:

Step(i):-

Given first sample size n₁ = 100

Given mean of the first sample x₁⁻ = 178

Standard deviation of the sample S₁ = 35

Given second sample size n₂= 100

Given mean of the second sample x₂⁻ = 193

Standard deviation of the sample S₂ = 37

Step(ii):-

Standard error of two population means

$se(x^{-} _{1} -x^{-} _{2} ) = \sqrt{\frac{s^{2} _{1} }{n_{1} }+\frac{s^{2} _{2} }{n_{2} } }$

$se(x^{-} _{1} -x^{-} _{2} ) = \sqrt{\frac{(35)^{2} }{100 }+\frac{(37)^{2} }{100 } }$

$se(x^{-} _{1} -x^{-} _{2} ) = 5.093$

Degrees of freedom

ν = n₁ +n₂ -2 = 100 +100 -2 = 198

t₀.₁₀ = 1.6526

Step(iii):-

90% confidence interval for the difference between the two population means

$(x^{-} _{1} - x^{-} _{2} - t_{\frac{\alpha }{2} } Se (x^{-} _{1} - x^{-} _{2}) , x^{-} _{1} - x^{-} _{2} + t_{\frac{\alpha }{2} } Se (x^{-} _{1} - x^{-} _{2})$

(178-193 - 1.6526 (5.093) , 178-193 + 1.6526 (5.093)

(-15-8.4166 , -15 + 8.4166)

( -23.4166 , -6.5834)

4 0

3 years ago

Solve this problem. What is n? 6n = 24 n=

dalvyx [7]

Answer:

Step-by-step explanation:

Divide 24 by 6, which equals 4

4 0

3 years ago

Read 2 more answers

ILEGIT NEED HELP BADDDDDDD

sergiy2304 [10]

Please restore my faith in humans and try to understand this concept

3 0

3 years ago

Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of grams of fat per pound

STatiana [176]

Complete question is:

Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviation of 7 grams of fat per pound. A random sample of 34 farm-raised trout is selected. The mean fat content for the sample is 29.7 grams per pound. Find the probability of observing a sample mean of 29.7 grams of fat per pound or less in a random sample of 34 farm-raised trout. Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

Answer:

Probability = 0.0277

Step-by-step explanation:

We are given;

Mean: μ = 32

Standard deviation;σ = 7

Random sample number; n = 34

To solve this question, we would use the equation z = (x - μ)/(σ/√n) to find the z value that corresponds to 29.7 grams of fat.

Thus;

z = (29.7 - 32)/(7/√34)

Thus, z = -2.3/1.200490096

z = -1.9159

From the standard z table and confirming with z-calculator, the probability is 0.0277

Thus, the probability to select 34 fish whose average grams of fat per pound is less than 29.7 = 0.0277

7 0

3 years ago