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3 years ago

Solve for x to the nearest

Mathematics

2 answers:

OlgaM077 [116]3 years ago

6 0

Answer:

in right angled triangleBCD

BC=√{DC²-BC²)=√{10²-6²)=8

again in right angled triangle ABC

AB=√(BC²-AC²)

x=√(8²-7²)=3.87

loris [4]3 years ago

3 0

Rad15
use the pythagorean theorem on both triangles to find x
y^2 + 6^2 = 10^2
y=8
7^2 + x^2 = 8^2
x=radical 15

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Five less than half the distance from Delaney’s home to the mall is more than 6 miles.

vovikov84 [41]

let x = the distance from Delany's house to the mall

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The legs of a right triangle XYZ are 5 cm and 12 cm. Triangle XYZ is dilated to triangle X'Y'Z' by a scale factor of 2/3 . What

allochka39001 [22]

Assuming that the base is 5 cm and the height is 12 cm, so by dilating the legs by a scale factor of 2/3, the new dimensions are:

base = 5 cm * (2 / 3) = 10/3 cm

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A = b h / 2

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3 years ago

A researcher wants to check the claim that convicted burglars spend less than 18.7 months in jail, on average. She takes a rando

vivado [14]

Answer:

a. No, there is not enough evidence to support the claim using a level of significance of 1%.

b. Null Hypothesis, $H_0$ : $\mu \geq$ 18.7 months

c. Alternate Hypothesis, $H_A$ : $\mu$ < 18.7 months

d. P-value of hypothesis test = 0.0199 or 1.99%

e. We conclude that the convicted burglars spend more than or equal to 18.7 months in jail, on average.

Step-by-step explanation:

We are given that a researcher wants to check the claim that convicted burglars spend less than 18.7 months in jail, on average.

She takes a random sample of 67 such cases from court files and finds that the mean is 16.8 months with a standard deviation of 7.3 months.

Let $\mu$ = true mean time convicted burglars spend in jail.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 18.7 months {means that the convicted burglars spend more than or equal to 18.7 months in jail, on average}

Alternate Hypothesis, $H_A$ : $\mu$ < 18.7 months {means that the convicted burglars spend less than 18.7 months in jail, on average}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean time spent in jail = 16.8 months

s = sample standard deviation = 7.3 months

n = sample of cases = 67

So, test statistics = $\frac{16.8-18.7}{\frac{7.3}{\sqrt{67} } }$ ~ $t_6_6$

= -2.13

Hence, the value of test statistics is -2.13.

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

P-value = P( $t_6_6$ < -2.13) = 0.0199 or 1.99%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.0199 or 1.99% which is clearly higher than the level of significance of 1%, so we will not reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the convicted burglars spend more than or equal to 18.7 months in jail, on average.

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daser333 [38]

Answer:

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Step-by-step explanation:

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