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3 years ago

HELP ME!!!! PLEASE!!!! Ughhh

Mathematics

2 answers:

professor190 [17]3 years ago

8 0

A 12/5. Opposite/Adjacent

galina1969 [7]3 years ago

5 0

Tan(α)= Opposite / Adjacent
Opposite =12
Adjacent = 5

tan(67°) = 12/5

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I need help with this please the answers I can choose from are

STatiana [176]

Sum of interior angles of a △ = 180°

∴ ∠T + ∠V + ∠U = 180°

⇒ 37° + 63° + X° = 180°

⇒ X° + 100 = 180°

⇒ X° = 180° - 100°

⇒ X° = 80° (D).

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3 years ago

No Links

Black_prince [1.1K]

Answer:

Step-by-step explanation:

A and B would be very expensive for 2 people, and D is too little, so C would be sensible! Hope this helps!

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2 years ago

What is 30% of a $24.50 item

Zarrin [17]

$24.50*0.3 = $7.35
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2 years ago

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IrinaVladis [17]

It is d

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2 years ago

A study of 4,430 U.S. adults found that 2,913 were obese or overweight. The 95% confidence interval for the proportion of adults

IrinaK [193]

Answer:

$0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644$

$0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672$

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

Step-by-step explanation:

We can begin founding the estimated proportion of people overweight:

$\hat p =\frac{2913}{4430}= 0.658$

We need to find a critical value for the confidence level using the normal standard distribution. We know that 95% is the confidence level, then the significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the true population proportion of interest is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing the data provided we got:

$0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644$

$0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672$

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

6 0

2 years ago