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4 years ago

I you were flipping a nickel a dime and a quarter how many possible outcomes are possible for getting heads or tails

Mathematics

1 answer:

zhuklara [117]4 years ago

8 0

3/6 possible out comes for heads and tails

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Find the P-value for the indicated hypothesis test with the given standardized test statistic, z. Decide whether to reject H0

Leviafan [203]

Answer: We do not reject the null hypothesis.

Step-by-step explanation:

When the p-value is greater than the significance level , then we do not reject the null hypothesis or if p-value is smaller than the significance level , then we reject the null hypothesis.

Given : Test statistic : $z = -2.28$

Significance level : $\alpha=0.02$

By using the standard normal distribution table ,

The p-value corresponds to the given test statistic ( two tailed ):-

$2P(Z>|z|)=2P(Z>2.28)=0.0226$

Since the p-value is greater than the significance level of 0.02.

Then , we do not reject the null hypothesis.

8 0

3 years ago

Find 250% of 60 please help thanks

____ [38]

Answer:

the answer is 150!

Step-by-step explanation:

250% of 60=150

4 0

3 years ago

Find the surface area for the following prisms.

liubo4ka [24]

Answer:

uixis9sugwhwo19wha

Step-by-step explanation:

isisisisisisi

7 0

3 years ago

X + 7 = 16<br> X = 16 - 7

vfiekz [6]

Answer:5

dhhddhhhhddkskkStep-by-step explanation:

8 0

2 years ago

Read 2 more answers

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

3 years ago