All categories

3 years ago

Can u help me with this question

Mathematics

1 answer:

marta [7]3 years ago

4 0

I can't see the full problem

You might be interested in

Pls help with this question please

Marina86 [1]

Answer:

$x=8$

Step-by-step explanation:

<h3>Given:-</h3>

$RS=28$

$RT=3x+4$

Solve for x.

<h3>According to question:-</h3>

[Tangent of same circle are equal]

$=>3x+4=28$

$=>3x=28-4$

$=>3x=24$

$=>x=\frac{24}{3}$

<h3>The value of x is 8.</h3>

5 0

2 years ago

Answer this question for brainliest

iragen [17]

Answer:

y = -7x + 26

I checked it on Desmos

7 0

2 years ago

Read 2 more answers

I need help with math!!! time is running out!!! will mark brainliest!!!

hoa [83]

1.

Answer B.

2.

Answer B.

3.

Answers A. and C.

3 0

2 years ago

"4. A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of de

SSSSS [86.1K]

Answer:

With a 0.01 significance level and samples of 50 and 40 cofee drinkers, there is enough statistical evidence to state that the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers.

The test is a one-tailed test.

Step-by-step explanation:

To solve this problem, we run a hypothesis test about the difference of population means.

$$$Sample mean $\bar X: \bar X=4.35$\\Sample mean $\bar Y: \bar Y=5.84$\\Population variance $\sigma^2_X: \sigma^2_X=1.2^2$\\Population variance $\sigma^2_Y: \sigma^2_Y=1.36^2$\\Sample size $n_X=50$\\Sample size $n_Y=40$\\Significance level $\alpha=0.01$\\Z criticals values (for a 0.01 significance)\\(Left tail test) Z_{1-\alpha}=Z_{0.990}=-2.32635\\($Right tail test) Z_{\alpha}=Z_{0.010}=2.32636\\($Two-tailed test) $Z_{1-\alpha/2}=Z_{0.995}=-2.57583$ and $ Z_{\alpha/2}=Z_{0.005}=2.57583\\\\$

The appropriate hypothesis system for this situation is:

$H_0: \mu_X-\mu_Y=0\\H_a: \mu_X-\mu_Y \leq 0\\\\$

Difference of means in the null hypothesis is:

$\mu_X - \mu_Y = M_0=0\\\\$

$$$The test statistic is$ Z=\frac{( \bar X-\bar Y)-M_0}{\sqrt{\frac{\sigma^2_X}{n_X}+\frac{\sigma^2_Y}{n_Y}}}\\$ $$$The calculated statistic is Z_c=\frac{[(4.35-5.84)-0]}{\sqrt{\frac{1.20^2}{50}+\frac{1.36^2}{40}}}=-5.43926\\p-value = P(Z \leq Z_c)=0.0000\\\\$

Since, the calculated statistic $Z_c$ is less than critical $Z_{1-\alpha}$ , the null hypothesis should be rejected. There is enough statistical evidence to state that the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers.

7 0

3 years ago

Read 2 more answers

The mathematics section of a standardized college entrance exam had a mean of 19.8 and an SD of 5.1 for a recent year. Assume th

Fudgin [204]

Answer:

a) 0.84% of students scored over 32

b) 29.12% of students scored under 17

c) 70.04% of students scored between 17 and 32.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 19.8, \sigma = 5.1$

a) About what percent of students scored over 32?

This is 1 subtracted by the pvalue of Z when X = 32. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{32 - 19.8}{5.1}$

$Z = 2.39$

$Z = 2.39$ has a pvalue of 0.9916

1 - 0.9916 = 0.0084

So 0.84% of students scored over 32

b) About what percent of students scored under 17?

This is the pvalue of Z when X = 17. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17 - 19.8}{5.1}$

$Z = -0.55$

$Z = -0.55$ has a pvalue of 0.2912

So 29.12% of students scored under 17

c) About what percent of students scored between 17 and 32?

This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 12. So

X = 32

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{32 - 19.8}{5.1}$

$Z = 2.39$

$Z = 2.39$ has a pvalue of 0.9916

X = 17

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17 - 19.8}{5.1}$

$Z = -0.55$

$Z = -0.55$ has a pvalue of 0.2912

0.9916 - 0.2912 = 0.7004

70.04% of students scored between 17 and 32.

5 0

2 years ago