Simplify 5a+7b-12a+6c-13b-22a+53c

Lostsunrise [7]

Answer:

the first option

4 0

3 years ago

In circle pp, ad¯¯¯¯¯ad¯ is a diameter and bc¯¯¯¯¯bc¯ is perpendicular to ad¯¯¯¯¯ad¯ at ee. if pd=7pd=7 and ae=4ae=4, find cece.

kvv77 [185]

See the attached figure.

<span>ad is a diameter of the circle with center p
</span>
∵ pd = radius = 7 ⇒⇒⇒ ∴ ad = 2 * radius = 2 * 7 = 14
∵ ae = 4 ⇒⇒⇒ ∴ ed = ad - ae = 14 - 4 = 10
∵ ad is a diameter
Δ acd is a triangle drawn in a half circle
∴ Δ acd is a right triangle at c
∵ bc ⊥ ad at point e
By applying euclid's theorem inside Δ acd
∴ ce² = ae * ed
∴ ce² = 4 * 10 = 40
∴ ce = √40 = 2√10 ≈ 6.325

3 0

4 years ago

One of the zookeepers solved to find Bernard’s speed when he is healthy using the equation:

MrRissso [65]

Answer:

Both the zoo-keepers are correct.

Step-by-step explanation:

The first zoo keeper uses the equation:

x - 41.5 = 13.5

⇒ x = 13.5 + 14. 5 = 55

Now, the second zoo- keeper uses the equation:

41.5 + 13.5 = x

⇒ x = 55

Both use x = 55.

4 0

3 years ago

Read 2 more answers

The Municipal Transit Authority wants to know if, on weekdays, more passengers ride the northbound blue line train towards the c

Kitty [74]

Answer:

The 90% confidence interval $-49.8$

The null hypothesis is $H_o : \mu_1 = \mu_2$

The alternative hypothesis $H_a : \mu_1 < \mu_2$

The distribution test statistics is $t = -3.222$

The rejection region is p-value < $\alpha$

The decision rule is reject the null hypothesis

The conclusion is

There is sufficient evidence to conclude that there are more passengers riding the 8:30 train

The p-value is $p-value =0.000951$

Step-by-step explanation:

From the question we are told that

The first sample size $n_1 = 30$

The first sample mean is $\= x _1 = 323$

The first standard deviation is $s_1 = 41$

The second sample size is $n_2 = 45$

The second sample mean is $\=x_2 = 356$

The second standard deviation is $s_2 = 45$

given that the confidence level is 90% then the level of significance is mathematically represented as

$\alpha = (100 -90)\%$

$\alpha = 0.10$

Generally the critical value of $\frac{\alpha }{2}$ obtained from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the pooled variance is mathematically represented as

$s^2 = \frac{(n_1 - 1)s_1^2 + (n_2 -1)s_2^2 }{n_1 + n_2 -2}$

$s^2 = \frac{(30 -1)(41^2) + (45-1)45^2}{30+45 -2}$

$s^2 = 1888.34$

Generally the standard error is mathematically represented as

$SE = \sqrt{\frac{s^2}{n_1} + \frac{s^2}{n_2} }$

=> $SE = \sqrt{\frac{1888.34}{30} + \frac{1888.34}{45} }$

=> $SE = 10.24$

Generally the margin of error is mathematically evaluated as

$E = Z_{\frac{\alpha }{2} } * SE$

$E = 1.645* 10.24$

$E = 16.85$

Generally the 90% confidence interval is mathematically represented as

$\=x_1 -\=x_2 -E < \mu_1 -\mu_2 < \=x_1 -\=x_2 +E$

$323 -356 -16.84$

$-49.8$

The null hypothesis is $H_o : \mu_1 = \mu_2$

The alternative hypothesis $H_a : \mu_1 < \mu_2$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \=x_2 }{SE}$

=> $t = \frac{323-356}{10.24}$

=> $t = -3.222$

Generally the degree of freedom is mathematically represented as

$df = n_1+n_2 -2$

$df = 30 + 45 -2$

$df = 73$

The p-value is obtained from the student t distribution table at degree of freedom of 73 at 0.05 level of significance

The value is $p-value =0.000951$

Here the level of significance is $\alpha = 5\% = 0.05$

Given that the p-value < $\alpha$ then we reject the null hypothesis

Then the conclusion is

There is sufficient evidence to conclude that there are more passengers riding the 8:30 train

5 0

3 years ago

F(x)=2-3x I need help with this please

Fynjy0 [20]

Answer:

F= 0.5

Step-by-step explanation:

5 0

3 years ago