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erastovalidia [21]
2 years ago
12

an ostrich egg weighs 2.9 pounds.The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds.write

an equation that could be used to find the weight of w, in pounds, of the emu egg
Mathematics
2 answers:
Alex73 [517]2 years ago
8 0

Answer:

2.9-w=1.6

Step-by-step explanation:

Let w be weight of emu egg in pounds.

We have been given that an ostrich egg weighs 2.9 pounds.The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds.

We can represent this information as:

2.9-w=1.6

Therefore, the equation 2.9-w=1.6 will give us the weight of the emu egg, w, in pounds.

Gennadij [26K]2 years ago
4 0

Answer:

egg

Step-by-step explanation:

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Could some one help me please / whats the answer for 4+3p= 19
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2 years ago
There are two major cell phone providers in the Colorado Springs, Colorado area, one called HTC and the other, Mountain Communic
Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

Hello!

The objective of this exercise is to compare the "Churn rate" between the two major cell phone providers in colorado springs.

For each provider, the number of subscribers was noted at the beginning and end of a month.

The study variables are:

X₁: Number of customers that cut ties with HTC over a month.

X₂: Number of customers that cut ties with Mountain Communications over a month.

To compare the churn rate of both companies, the parameters of interest are the population proportions of customers that stopped using the companies services over the given period.

The hypotheses of interest are:

H₀: p₁ = p₂

H₁: p₁ ≠ p₂

α: 0.01

The statistic to use is the standard normal approximation:

Z= \frac{('p_1-'p_2)- (p_1-p_2)}{\sqrt{'p(1 - 'p)[\frac{1}{n_1} + \frac{1}{n_2}  ]} } }

Z≈N(0;1)

HTC

n₁= 10000 customers (beginning of the month)

x₁= 190 customers (people that stopped using the company service)

sample proportion 'p₁= x₁/n₁= 190/10000= 0.019

Mountain Communications

n₂= 12285 customers (beginning of the month)

x₂= 215 customers (people that stopped using the company service)

sample proportion 'p₂= x₂/n₂= 215/12285= 0.0175

pooled sample proportion 'p= \frac{x_1 + x_2}{n_1 + n_2} = \frac{190 + 215}{10000 + 12285}

'p= 0.0182

Z_{H_0}= \frac{(0.019 - 0.0175) - 0}{\sqrt{(0.0182*0.9818)[\frac{1}{10000} + \frac{1}{12285}  ]} }

Z_{H_0}= 0.833

The p-value for this test is: 0 .4048

Note: Using the p-value method to decide over a hypothesis test, the decision rule is:

If p-value ≥ α, then you do not reject the null hypothesis.

If p-value < α, then you reject the null hypothesis.

The p-value < α, so the decision is to reject the null hypothesis.

With a significance level of 1%, there is enough evidence to conclude that there is a difference between the churn rate of the two cell phone providers.

I hope it helps!

3 0
2 years ago
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