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2 years ago

What is the measure of A 54° D. 66° B. 56° E. 114° C. 58°

Mathematics

2 answers:

geniusboy [140]2 years ago

7 0

I’m pretty sure it’s 66 Becuase whatever u do to one side u do the other ‍♂️

dangina [55]2 years ago

5 0

Answer:

66 I think might he wrong

Step-by-step explanation:

I'm not sure if its right

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At the city Museum child admission is $6.40 in adult admission is $9.70 on Sunday 149 tickets were sold for a total sales of $11

Law Incorporation [45]

Answer:

Adult=58

Step-by-step explanation:

c=child, a=adult

6.4c+9.7a=1145 equation 1

c+a=149 equation 2

a=149-c modified equation 2 to isolate a

6.4c+9.7(149-c)=1145 substitute value of a from equation 1 into equation 2

6.4c+1445.3-9.7c=1145

-3.3c=-300.3

c=91

solve for a

c+a=149

91+a=149

a=58

Check answer:

6.4c+9.7a=1145

6.4(91)+9.7(58)=1145

582.40+562.60=1145

1145=1145

8 0

2 years ago

franklin is thinking of two numbers. the sum of the two numbers is 40.the first number x, is 11 less than twice the second numbe

vampirchik [111]

Answer: d) x+y =40

x=2y -11

Step-by-step explanation:

5 0

2 years ago

keeping in mind the role the order of precedence plays in equations, what would excel display as the result of the following equ

Alex73 [517]

The answer to youre question is 5.5

8 0

3 years ago

3.14 times 16 divided by 3

AleksAgata [21]

Answer:

The answer is 16.746667

Step-by-step explanation:

1) Simplify 3.14 × 16 to 50.24.

$50.24 \div 3$

2) Simplify 50.24 ÷ 3 to 16.746667.

$16.746667$

Therefor. the answer is 16.746667.

5 0

2 years ago

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.

8 0

2 years ago