What is the perimeter of the shaded figure?

Given a minimum usual value of 135.8 and a maximum usual value of 155.9, determine which (1 point) of the following values would

maxonik [38]

Answer: b. 134

Step-by-step explanation:

Given : A minimum usual value of 135.8 and a maximum usual value of 155.9.

Let x denotes a usual value.

i.e. 135.8< x < 155.9

Therefore , the interval for the usual values is [135.8, 155.9] .

If interval for any usual value is [135.8, 155.9] , then any value should lie in this otherwise we call it unusual.

Let's check all options

a. 137 ,

since 135.8< 137 < 155.9

So , it is usual.

b. 134

since 134<135.8 (Minimum value)

So , it is unusual.

c. 146

since 135.8< 146 < 155.9

So , it is usual.

d. 155

since 135.8< 1155 < 155.9

So , it is usual.

Hence, the correct answer is b. 134 .

4 0

3 years ago

Convert the Cartesian equation (x 2 + y 2)2 = 4(x 2 - y 2) to a polar equation.

SashulF [63]

ANSWER

${r}^{2} = 4 \cos2\theta$

EXPLANATION

The Cartesian equation is

${( {x}^{2} + {y}^{2} )}^{2} = 4( {x}^{2} - {y}^{2} )$

We substitute

$x = r \cos( \theta)$

$y = r \sin( \theta)$

and

${x}^{2} + {y}^{2} = {r}^{2}$

This implies that

${( {r}^{2} )}^{2} = 4(( { r \cos\theta) }^{2} - {(r \sin\theta) }^{2} )$

Let us evaluate the exponents to get:

${r}^{4} = 4({ {r}^{2} \cos^{2}\theta } - {r}^{2} \sin^{2}\theta)$

Factor the RHS to get:

${r}^{4} = 4{r}^{2} ({ \cos^{2}\theta } - \sin^{2}\theta)$

Divide through by r²

${r}^{2} = 4 ({ \cos^{2}\theta } - \sin^{2}\theta)$

Apply the double angle identity

$\cos^{2}\theta -\sin^{2}\theta= \cos(2 \theta)$

The polar equation then becomes:

${r}^{2} = 4 \cos2\theta$

7 0

3 years ago

Sunspots have been observed for many centuries. Records of sunspots from ancient Persian and Chinese astronomers go back thousan

Lunna [17]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41

Alternate Hypothesis, $H_A$ : $\mu$ > 41

(b) The value of z test statistics is 1.08.

(c) We conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

Step-by-step explanation:

We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.

It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.

Let $\mu$ = mean sunspot activity during the Spanish colonial period.

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41 {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}

Alternate Hypothesis, $H_A$ : $\mu$ > 41 {means that the mean sunspot activity during the Spanish colonial period was higher than 41}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean = 47

σ = population standard deviation = 35

n = sample of periods from Spanish colonial times = 40

So, the test statistics = $\frac{47-41}{\frac{35}{\sqrt{40} } }$

= 1.08

(b) The value of z test statistics is 1.08.

(c) Now, the P-value of the test statistics is given by;

P-value = P(Z > 1.08) = 1 - P(Z < 1.08)

= 1 - 0.8599 = 0.1401

Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

8 0

3 years ago

Giant is selling Doritos 2 family bags for 7.00 I’m buying them for the class to celebrate great test scores if I buy 13 bags ho

elena55 [62]

$45.50
2 bags = $7
13 bags=?
7÷2=3.5
3.50=1 bag
$3.50 × 13 bags = $45.50 for 13 bags of Doritos

6 0

3 years ago

Equivalent fraction for ½.

labwork [276]

Answer:

2/4

Step-by-step explanation:

half/whole for any number really

5 0

3 years ago