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3 years ago

Plz halp wil giv brainlyes

Mathematics

1 answer:

LuckyWell [14K]3 years ago

6 0

Answer:

∠F = 36.1°

Step-by-step explanation:

tan(x) = $\frac{O}{A} = \frac{3.5}{4.8}$

x = $tan^{-1} (\frac{3.5}{4.8})$ = 36.1°

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The length of a rectangle is represented by the function L(x) = 5x. The width of that same rectangle is represented by the funct

Mazyrski [523]

Remember that the area of a rectangle is the length of the rectangle multiplied by the width of the rectangle.

In this case, we could say (where $A(x)$ is the area of the rectangle):

$A(x) = L(x) \cdot W(x)$

Substituting the values the problem gave us for $L(x)$ and $W(x)$ , we can find the formula for $A$ in terms of $x$ , which is:

$A(x) = (5x) \cdot (2x^2 - 4x + 13) = (10x^3 - 20x^2 + 65x)$

The formula for the area of the rectangle would be A(x) = 10x³ - 20x² + 65x.

3 0

3 years ago

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Hi can you guys help me with this please

Sav [38]

Answer:

Step-by-step explanation:

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4 0

2 years ago

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Find m angle dbc in the figur below

Gemiola [76]

Answer:

It like the once like Line with smaller angle (DBC) measure wouldn't 30* and by the exam the ratio of angle use by subjective by multiplying as large of angle.

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3 0

2 years ago

It is believed that 36% of the US population has never been married, 32% are divorced, 27% are married, and 5% are widowed. You

Katena32 [7]

Answer: the smallest number of people required for the sample to meet the conditions for performing inference is 100

Step-by-step explanation:

Given that;

36% of US population has never been married

32% are divorced

27% are married

5% are widowed

Taking a simple random sample of individuals to test this claim;

we need expected count in each cell to be at least 5, here the smallest proportion is 5% = 0.05

so we only need to satisfy condition for its expected count;

n × 0.05 ≥ 5

n = 5 / 0.05 = 100

Therefore the smallest number of people required for the sample to meet the conditions for performing inference is 100

6 0

2 years ago

Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams

Nataliya [291]

Answer:

a) 0.2318

b) 0.2609

c) No it is not unusual for a broiler to weigh more than 1610 grams

Step-by-step explanation:

We solve using z score formula

z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.

(a) What proportion of broilers weigh between 1150 and 1308 grams?

For 1150 grams

z = 1150 - 1387/192

= -1.23438

Probability value from Z-Table:

P(x = 1150) = 0.10853

For 1308 grams

z = 1308 - 1387/192

= -0.41146

Probability value from Z-Table:

P(x = 1308) = 0.34037

Proportion of broilers weigh between 1150 and 1308 grams is:

P(x = 1308) - P(x = 1150)

0.34037 - 0.10853

= 0.23184

≈ 0.2318

(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?

1510 - 1387/192

= 0.64063

Probabilty value from Z-Table:

P(x<1510) = 0.73912

P(x>1510) = 1 - P(x<1510) = 0.26088

≈ 0.2609

1610- 1387/192

= 1.16146

Probability value from Z-Table:

P(x<1610) = 0.87727

P(x>1610) = 1 - P(x<1610) = 0.12273

≈ 0.1227

No it is not unusual for a broiler to weigh more than 1610 grams

8 0

3 years ago